Chapter 13 Unsaturated...

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.

a. b.C CH

CH

HO H C C

HC C

H

H

H

HC

C

HH

HH

HH

HH

C HH

HCH2 CHCH2OH (CH3)2C CH(CH2)2CH3= =

c. C C C CH

CH

C

C

CH HH

HH

HH

HH

CHH

H C H

H

H

H(CH3)2CHC CCH2C(CH3)3 =

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.

a. CH2 CHCHCH2CH3

CH3

Answer: 3-methyl-1-pentene

5 C's in the longest chainpentene

CH2 CHCHCH2CH3

CH3

1 2 3 4 5

methyl at C3double bond at C1

��E\�0F*UDZ�+LOO�(GXFDWLRQ��7KLV�LV�SURSULHWDU\�PDWHULDO�VROHO\�IRU�DXWKRUL]HG�LQVWUXFWRU�XVH��1RW�DXWKRUL]HG�IRU�VDOH�RU�GLVWULEXWLRQ�LQ�DQ\�PDQQHU��7KLV�GRFXPHQW�PD\�QRW�EH�FRSLHG��VFDQQHG��GXSOLFDWHG��IRUZDUGHG��GLVWULEXWHG��RU�SRVWHG�RQ�D�ZHEVLWH��LQ�ZKROH�RU�SDUW�

Unsaturated Hydrocarbons 13–2

b. (CH3CH2)2C CHCH2CH2CH3

Answer: 3-ethyl-3-hepteneCH3CH2C CHCH2CH2CH3

CH2CH3

7 C's in the longest chainheptene

CH3CH2C CHCH2CH2CH3

CH2CH31 23

4 5 6 7

ethyl at C3

double bond at C3

CH3CH2CH CHCH=CHCH3c. Answer: 2,4-heptadiene

7 C's in the longest chainheptadiene

CH3CH2CH CHCH=CHCH3

1234567

double bonds at C2 and C4

CH2CH3d. Answer: 3-ethylcyclopentene

5 C's in the ringcyclopentene ethyl at C3

1 23

4

5 CH2CH3


[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.

12345

a. CH3CH2CH2CH2CH2C CCH(CH3)2

CH3CH2CH2CH2CH2C CCHCH3

CH3

Answer: 2-methyl-3-nonyneCH3CH2CH2CH2CH2C CCHCH3

CH3

9 C's in the longest chainnonyne triple bond at C3

b. C C CH2 CCH2CH3

CH3

CH3

CH3CH2 Answer: 6,6-dimethyl-3-octyne1 2 3 4 5

6

C C CH2 CCH2CH3

CH3

CH3

CH3CH2

7 8

triple bond at C32 methyl groups at C68 C's in the longest chain

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.

a. 4-methyl-1-hexene CH2 CHCH2CHCH2CH3

CH36 carbon chain double bond at C1

C C C C C C1 2 3 4 5 6

CH3

methyl at C4


Chapter 13–3

CHCH2CHCH2CH3

CH2CH3

CH3C

CH3

b. 5-ethyl-2-methyl-2-heptene

7 carbon chain double bond at C2

C C C C C CC1 2 3 4 5 6 7

CH3 CH2CH3

ethyl at C5methyl at C2

CH3CHC CCHCH3

CH3

CH3

c. 2,5-dimethyl-3-hexyne6 carbon chain

triple bond at C3

1 2 3 4 5 6C C C C C C

CH3 CH3

methyl at C2 methyl at C5

CH2CH2CH3

!d. 1-propylcyclobutene

4 carbon ringdouble bond at C1

CC C

C1

23

4CH2CH2CH3

propyl at C1

e. 1,3-cyclohexadiene

12

34

5

6

6 carbon ring2 double bonds (C1 and C3)

CH3CH2CH2CH2CH2CH2CHCH2C CH

CH2CH3

f. 4-ethyl-1-decyne12345678

10 carbon chain triple bond at C1

CCCCCCCCCCCH2CH3

910

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.

C CH

CH3

H

CH2CH2CH2CH2CH3

CH3CH CHCH2CH2CH2CH2CH3

1 2 3 4 5 6

a. cis-2-octene8 carbon chain 7 8

cis isomerboth alkyl groups on the same side

C CCH2CH2CH3

CH3CH2

H

Hb. trans-3-heptene

7 carbon chainCH3CH2CH CHCH2CH2CH3

1 2 3 4 5 6 7

trans isomeralkyl groups on opposite sides



c. trans-4-methyl-2-pentene C CCHCH3

CH3

H

H

CH3

5 carbon chain

CH3CH CHCH(CH3)CH3

1 2 3 4 5

trans isomeralkyl 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.

CH2 CHCH2CH2CH3

CH3CH2CH CHCH3a.

b.

c. CH3 C CHCH2CH2CH3

CH3

Each C has one H and one alkyl group.Cis and trans isomers are possible.

two H'scannot have cis or trans isomers

two CH3'scannot 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.

C CC

HOCH2(CH2)7CH2

H

H

CH H

CH2CH2CH3

trans

cis

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.

anda. CH3CH CHCH2CH3 CH2 CHCH2CH2CH3

C bonded to one H, one CH3 C bonded to two H'sdifferent connectivity

constitutional isomers

C CH

CH3CH2 CH3

HC C

H

CH3CH2 H

CH3

C CCH3

CH3CH2 H

HC C

H

CH3CH2 CH3

H

and

and

b.

c.

cis isomer trans isomersame connectivity

different 3-D arrangementstereoisomers

C bonded to one CH2CH3 and one CH3

C bonded to one H and one CH2CH3

different connectivityconstitutional isomers


Chapter 13–5

13.11 Double bonds in naturally occurring fatty acids are cis.

C CCH2

H H

CH2

CCH H

CH2

CCH

CH3CH2CH2CH2CH2

HC C

H H

CH2CH2CH2COHO

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 acidno double bonds

higher melting point63 °C

palmitoleic acidone double bond

lower melting point1 °C

13.13 The functional groups in tamoxifen are labeled.

C C

OCH2CH2N(CH3)2

CH3CH2

amineether

alkene

aromatic ring

aromatic ringaromatic ring

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

CH3CH2

O

OHC CH

levonorgestrel

O

OHC C CH3

CH3

(CH3)2N

RU 486

amine aromatic ring hydroxyl

alkyne

alkene

ketone

alkene

ketone

alkene

hydroxyl

alkyne



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.

a. CH3CH2CH CHCH2CH3

H2Pd

CH3CH2CH2CH2CH2CH3

C CCH3

CH3 CH2CH(CH3)2

H

b.H2Pd

CH3CHCH2CH2CH(CH3)2

CH3

CH3c. H2

Pd

CH3

13.16 To draw the product of each halogenation reaction, add a halogen to both carbons of the double

bond.

a. CH3CH2CH CH2 Cl2+ CH3CH2C

Cl

C H

H

Cl

H

Br2+b.CH3

CH3

CH3

CH3

Br

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.

CH3CH CHCH3a. + HBr CH3C CCH3

H

HBr

H

CH3

b. + HBrCH3

HH

Br

This C has more H's.Add H here.

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

(CH3)2C CHCH3c. + HCl C CCH3

H

H

ClCH3

CH3

This C has more H's.Add H here.

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


Chapter 13–7

d. + HCl ClH

HH

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.

CH3CH CHCH3a. CH3C CCH3

HO H

H HH2SO4

H OH

CH3CH2CH CH2b. CH3CH2CHO

CH

HH

HThis C has 2 H's so the H bonds here.

This C has one H so the OH bonds here.

H2SO4

H OH

CH3

CH3

c. CH3CH3

OH

HH2SO4

H OH

13.19 Draw the products of each reaction.

PdH2 CH3CH2CH2C

H

HC HH

Ha. CH3CH2CH2CH CH2

CH3CH2CH2CH

ClC HCl

Hb.

Cl2CH3CH2CH2CH CH2

CH3CH2CH2CBr

HC HH

Brc.

Br2CH3CH2CH2CH CH2

CH3CH2CH2CH

BrC HH

Hd. H Br

CH3CH2CH2CH CH2

CH3CH2CH2CH

ClC HH

He.

H ClCH3CH2CH2CH CH2

CH3CH2CH2CHO

HC HH

Hf. H2SO4

H OHCH3CH2CH2CH CH2



13.20 Draw the products of the hydrogenation reactions.

C CCH2

H H

CH2CH2CH2CH2CH2CH2CH2COOHCC

H

CH3CH2

H

CH2CH2CH2

CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2COOHCC

H H

CH2CC

H

CH3CH2

H

a.C C

CH2

H H


H H

CH2CC

H

CH3CH2

H

H2Pd

CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2COOHCH3CH2CH2CH2CH2CH2CH2b. 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.

CH

HC C

HC

HC C

CH3HCH3

CH2 CH2

CH3

H CH2

CH3 CH3 CH3

CH2 CCH3

CH2CH3

a. CH2 CCH3

CH2CH3

CH2 CCH3

CH2CH3

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

C CH

H

CH3

CNCH

HCCH3

CNCH

HCCH3

CN

CH2 CCH3

CNb. CH2 C

CH3

CNCH2 C

CH3

CN


CH

CH

HCH

CH

HCH

CH

H

Cl Cl Cl

CH2 CH

Cl

c. CH2 CH

Cl

CH2 CH

Cl



Chapter 13–9

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

CH2 CH

OCOCH3

poly(vinyl acetate)

CH2C CH2C CH2C

O O O

H H H

COCH3 COCH3 COCH3

formed from

Break these bonds to form the monomer.

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.

CH2CH3

I

OH

CH2CH2CH2CH3

CH3Br

Cl

a.

b.

c.

d.

CH2CH2CH3 propyl

propylbenzene

ethyl

iodop-ethyliodobenzene

butyl

OH on benzene ring = phenol

m-butylphenol

bromo

chloro

CH3 on benzene ring = toluene

2-bromo-5-chlorotoluene

13.24 Draw the structure corresponding to each name.

!!!d. 4-chloro-1,2-diethylbenzene

CH2CH2CH2CH2CH3

ClCl

NH2

Br

CH2CH3

CH2CH3

Cl

a. pentylbenzene

b. o-dichlorobenzene

!!!c. m-bromoaniline

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.



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

Cl

NO2

Cl

ClSO3H

a.

b.

c.

Cl

Cl

Cl

Cl

Cl

Cl

Cl2FeCl3

HNO3

SO3H2SO4

H2SO4

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.

Cl2FeCl3

CH3 CH3 CH3Cl

ClCl

o-chlorotoluene m-chlorotoluene p-chlorotoluene

CH3

Solutions to End-of-Chapter Problems 13.29

OCH3

CH3anethole

a. molecular formula: C10H12Ob. aromatic ring, alkene, etherc. trans d. Tetrahedral C's are indicated. All other C's are trigonal planar.

trans alkene

aromatic ringether

tetrahedral

tetrahedral

13.30

methyl cinnamate

a. molecular formula: C10H12O2b. aromatic ring, esterc. trans d. Tetrahedral C is indicated. All other C's are trigonal planar.

trans alkene

aromatic ring

tetrahedral

O

O

CH3


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.

HC CCH2CH2CH2CH3a. b. CHCH2CH2CH CH2CH2 c.

13.33 Draw three alkynes with molecular formula C5H8.

HC CCH2CH2CH3 CH3C CCH2CH3 HC C CCH3

CH3

H

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

CHCH2CH2CH3CH2 CH3CH CHCH2CH3 CCH2CH3CH2CH3

CH3C CHCH3CH3

CHCHCH3CH2CH3

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

C CHb.

c.a.tetrahedral

linear

trigonal planar

trigonal planar

all trigonal planar

trigonal planar

tetrahedral

tetrahedral

13.36 Predict the indicated bond angles in falcarinol.

C CHH H

COH

HC C C C CH2 C

HC (CH2)6CH3H

a b c

d e

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



13.37 Give the IUPAC name for each compound.

a. b. 1 2 3 4

5

1234

2-ethyl

4 C chainbutene 2-ethyl-1-butene

6

6 C chainhexene 2-hexyne

13.38 Give the IUPAC name for each compound.

HC C CH2CH2CH2CH2CH3a.

7 C chain 1- heptyne

b.CH2CH3

2-ethylcyclopentene5 carbon ring; ethyl at C2

12



double bond at C1

a. CH2 CHCH2CH2C(CH3)3

Answer: 5,5-dimethyl-1-hexeneCH2 CHCH2CH2CCH3

CH3

CH3

6 C's in the longest chainhexene

CH2 CHCH2CH2CCH3

CH3

CH3

1 2 3 45

2 methyls at C5

b. (CH3CH2)2C CHCHCH2CHCH3

CH3 CH3

3-ethyl-5,7-dimethyl-3-octene8 C's in the longest chainoctene

1 2 3 4 5

2 methyls at C5 and C7

CH3CH2C CHCHCH2CHCH3

CH3 CH3CH3CH2

CH3CH2C CHCHCH2CHCH3

CH3 CH3CH3CH2

6 7 8

double bond at C3

ethyl at C3

Answer:

c. CH2 C CH2CH3

CH2CH2CH2CH2CH3

Answer: 2-ethyl-1-heptene


1 2

3 4 5 6 7double bond at C1

CH2 C CH2CH3

CH2CH2CH2CH2CH3

ethyl at C2


Chapter 13–13

CH3C CCH2C(CH3)3d.

Answer: 5,5-dimethyl-2-hexyne

66 C's in the longest chain

hexyne

1 2 35

CH3C CCH2CCH3

CH3

CH3

CH3C CCH2CCH3

CH3

CH3

2 methyls at C5triple bond at C2

CH3C C CH2 CHCH3

CCH3

CH2CH3

CH2CH3

e. Answer: 6-ethyl-5,6-dimethyl-2-octyne

68

8 C's in the longest chainoctyne

1 2 3 45


CH3C C CH2 CHCH3

CCH3

CH2CH3

CH2CH3triple bond at C2

ethyl at C6

CH2 CHCH2 CCH3

CH3

CH=CH2f. Answer: 3,3-dimethyl-1,5-hexadiene

1


6 C's in the longest chainhexadiene

6 5 43

2 methyls at C3

CH2 CHCH2 CCH3

CH3

CH=CH2



double bond at C1

a. CH2 CHCH2CHCH2CH3

Answer: 4-methyl-1-hexene

6 C's in the longest chainhexene

1 2 3 4 5

methyl at C4

CH3

CH2 CHCH2CHCH2CH3

CH3

CH2 CHCH2CHCH2CH3

CH3

6

b.

C CHCH2CHCH2CH3 Answer: 5-ethyl-2-methyl-2-octene

8 C's in the longest chainoctene

CH2CH2CH3

(CH3)2C CHCH2CHCH2CH3

CH2CH2CH31 2 3

8

5

methyl at C26

CH3

CH3

C CHCH2CHCH2CH3

CH2CH2CH3

CH3

CH3

double bond at C2ethyl at C5



c.

Answer: 4-methyl-2-heptyne

7 C's in the longest chainheptyne

1 2 4

75

triple bond at C2methyl at C4

CH3C CCHCH3

CH2CH2CH3

CH3C CCHCH3

CH2CH2CH3

CH3C CCHCH3

CH2CH2CH3

d.

Answer: 2,2,5,5-tetramethyl-3-hexyne

6 C's in the longest chainhexyne

36

triple bond at C3


(CH3)3CC CC(CH3)3

CH3CC CCCH3

CH3

CH3 CH3

CH3

CH3CC CCCH3

CH3

CH3 CH3

CH352

e.

Answer: 3-butyl-2-heptene


double bond at C2

butyl at C3

1

(CH3CH2CH2CH2)2C CHCH3

CH3CH2CH2CH2C CHCH3

CH2CH2CH2CH3

CH3CH2CH2CH2C CHCH3

CH2CH2CH2CH3

37

f.

Answer: 2,8-dimethyl-2,7-nonadiene

9 C's in the longest chainnonene


methyls at C2 and C8

1

(CH3)2C

2 7

CHCH2CH2CH2CH C(CH3)2

CH3C CHCH2CH2CH2CH CCH3

CH3CH3

CH3C CHCH2CH2CH2CH CCH3

CH3CH3

93 8

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

CH3

CH2CH3

CH2CH3a. b.4-methylcyclohexene 3,3-diethylcyclobutene

6 carbon ringcyclohexene

methyl at C4

double bond at C1

4 carbon ringcyclobutene

2 ethyl groups at C3

1

2 34

1

2 3

4


Chapter 13–15

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

a. b.4-propylcylcopentene 1,4-cycloheptadiene

5 carbon ringcyclopentene

propyl at C4

double bond at C1

1

23

4

51

2 34

CH2CH2CH3


7 carbon ringcycloheptene

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.

CH2 CHCHCH2CH2CH2CH2CH3

CH38 carbon chain double bond at C1 methyl at C3

a. 3-methyl-1-octene C C C C C C1 2 3 4 5 6

C C7 8

CH3

CH2CH3

b. 1-ethylcyclobutene CC C

CCH2CH3

4 carbon ringdouble bond at C1

ethyl at C1

methyl at C2

c. 2-methyl-3-hexyne C C C C C C1 2 3 4 5 6

CH3

CH3 CH

CH3

C CCH2CH3

6 carbon chaintriple bond at C3

d. 3,5-diethyl-2-methyl-3-heptene

2 ethyl groups at C3 and C5

7 carbon chaindouble bond at C3

C C C C C C1 2 4 5 6

C7

CH3

methyl at C2

CH2CH3

CH2CH3

CH3 CCH3

C C CCH2CH3

CH2CH3

HCH2CH3

H H

3

!e. 1,3-heptadiene C C C C C C1 2 3 4 5 6

C7

7 carbon chaindouble bonds at

C1 and C3

CH2 C C CHCH2CH2CH3HH

C CCH3

H

CH2CH2CH2CHCH3

H CH38 carbon chain

double bond at C2

f. cis-7-methyl-2-octene C C C C C C1 2 3 4 5 6

C C7 8

methyl at C7

CH3

cis



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.

5 carbon ring double bond at C1

methyls at C1 and C2a. 1,2-dimethylcyclopentene1

2

CH3

CH3

1

2

CH3

CH3


b. 6-ethyl-2-octyne

ethyl at C6

C C C C C1 2 3 4 5 6

C C CCH2CH3

7 8C CCH2CH2CHCH2CH3CH3

CH2CH3

5 carbon chain double bonds at C1 and C4

c. 3,3-dimethyl-1,4-pentadiene

2 methyls at C3

C C C C C1 2 4 5

CH CCH2

CH3

CH3

CH3

CH3

CH CH2

6 carbon chain double bond at C2

d. trans-5-methyl-2-hexene

methyl at C5

C C C C1 2 4 5

C CC CCH3

3 6 CH3 HCH2CHCH3H

CH3trans


e. 5,6-dimethyl-2-heptyne

methyls at C5 and C6

C C C C1 2 4 5

C CCH3

3 6C

CH3

7C CCH3 CH2 CH CHCH3

CH3 CH3


f. 3,4,5,6-tetramethyl-1-decyne

methyls at C3, C4, C5 and C6

C C C C1 2 4 5

C CCH3

3 6C

CH3

7CH C CH CH CH

CH3 CH3CH3 CH3

C C C8 9 10

CH3

CH(CH2)3CH3

CH3

13.45 Correct each of the incorrect IUPAC names.

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

CH3C CHCH2CH2CH3

CH3

2-methyl-2-hexene

b. The name 1-methylbutene makes the last carbon in the chain a substituent. In addition, the location of the double bond is not specified. There are five carbons in the chain (not a methyl substituent): 2-pentene.

2-pentene

CH3CH CHCH2CH3

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

CH3

CH3

12

3

45

6


Chapter 13–17

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

3-methyl-1-heptyne

1 23

HC CCCH3CH2CH2CH2CH3

H

13.46 Correct each of the incorrect IUPAC names.

a. The alkene has two methyl groups off of C2. Therefore, it cannot be cis.

CH3C CHCH2CH2CH3

CH3

2-methyl-2-hexene

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

C CHCHCH3CH3

CH2

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

CH3 12

3

456CH3

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

3,3-dimethylcyclohexene

12 3

456

CH3CH3

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.

CC

CC

CC

CHCHCH2CH2CH2COOHCH2 S CH2

CHCONHCH2COOHNHCOCH2CH2CHCOOH

NH2

HH

H

H

CCCH3CH2CH2CH2CH2

H

H

H H

OHcis

trans



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

C C(CH2)12CH3

H

CH3(CH2)8CH

H

cisa. b. C C

CH

H

OH

trans

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.

1

2

3

4

5

4-methylcis

cis-4-methyl-2-pentene

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.

H3C

CH3

CH3 trans-2-methyl-3-heptene

2-methyltrans

1 2 3

4

5

6 7

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

C CCH3 CH2CH2CH2CH2CH2CH3

HH

C CCH3 H

CH2CH2CH2CH2CH2CH3H

C CCH3CH CH2CH2CH3

HH

C CCH3CH H

CH2CH2CH3H

cis-2-nonene

trans-2-nonene

CH3

CH3

cis-2-methyl-3-heptene

trans-2-methyl-3-heptene

a. b.


Chapter 13–19

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

C CCH3CH2 CH2CH2CH3

HH

C CCH3CH2 H

CH2CH2CH3H

C CCH3 C

HH

C CCH3 H

H

cis-3-heptene

trans-3-heptene

cis-4,4-dimethyl-2-hexene

trans-4,4-dimethyl-2-hexene

a. b.CH3

CH3

CH2CH3

CCH3

CH3

CH2CH3

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.

C C

CH3 H

H CH

CCH3

H C CCH3 H

H CH

CH

CH3 C CH H

CH3 CH

CH

CH3

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

a.

and

same molecular formulasame connectivity

identical

andb.

same molecular formuladifferent connectivity

constitutional isomers



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

CH3anda.


identical


different arrangement in space(cis and trans isomers)

stereoisomers

CH3

H3C

CH3

CH3 H3C CH3

CH3

b. and

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

CH2 CHCH2CH2CH2CH3

CH2

CH3

(CH3)2C CHCH2CH2CH3

a.

b.

c.

d.

CH3CH2CH2CH2CH2CH3

CH3

CH3

(CH3)2CHCH2CH2CH2CH3 H

H2Pd

H2Pd

H2Pd

H2Pd

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

CH2 CHCH2CH2CH2CH3

CH2

CH3

(CH3)2C CHCH2CH2CH3

a.

b.

c.

d.

CH2CHCH2CH2CH2CH3

CH2Br

CH3

(CH3)2C Br

Br2

Br2

Br2

Br2

Br Br

BrCHCH2CH2CH3Br

Br

Br

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

(CH3)2C C(CH3)2

CH3CHCH2CH(CH3)2

CH3

Cl

HCl

Cl

Cl

(CH3)2C C(CH3)2

CH2 CHCH2CH(CH3)2

CH2d.

c.

b.

a. HCl

HCl

HCl

HCl

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

(CH3)2C C(CH3)2

CH3CHCH2CH(CH3)2

CH3

OH

HOH

OH

OH

(CH3)2C C(CH3)2

CH2 CHCH2CH(CH3)2

CH2d.

c.

b.

a. H2SO4

H2SO4

H2SO4

H2SO4+ H2O

+ H2O

+ H2O

+ H2O


Chapter 13–21

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

CH2CH3

CH2CH3

CH2CH3

CH2CH3H

HH

ClH

Cl

CH2CH3BrH

Br

Cl

HH

CH2CH3Br

HH

HH

OH

a.

b.

c.

d.

e.

f.

CH2CH3 H2Pd

CH2CH3 Cl2

CH2CH3 Br2

CH2CH3 HCl

CH2CH3 HBr

CH2CH3 H2OH2SO4

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

b.

d.

e.

Cl2

+ H2Pd

CH3

CH3 CH3Cl

ClCH3

c.

CH3CH CHCH3 + HCl CH3CHCH2CH3

Cl

CH3CH CHCH2CH2CH3 + Br2 CH3CHCHCH2CH2CH3

BrBr

(CH3)2C CHCH2CH3 + HBr (CH3)2CCH2CH2CH3Br

f.CH3

CH3

CH3+ H2O

CH3

CH3

CH3

OH

H2SO4

13.63 Work backwards to determine what alkene is needed as a starting material to prepare each of the

alkyl halides or dihalides.

CH3CH2Br

Cl

Cl

ClCH3

BrCH2CHCH2CH(CH3)2Br

a.

b.

c.

d.

CH2 CH2 CH3

CH2 CHCH2CH(CH3)2

HBr

HCl

Cl2

Br2

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.

CHCH2CH3CH2

1-butene

+ HBr CH3CHCH2CH3

Br2-bromobutane

CH3CH CHCH3 + HBr CH3CHCH2CH3

Br2-bromobutane2-butene

CHCH2CH2CH3CH2 + HBr CH3CHCH2CH2CH3

Br2-bromopentane1-pentene



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

HBrHCl

Cl2

Br2

a.

b.

c.

d.

e.

f.

(CH3)3CCl

(CH3)3CH (CH3)2CCH2BrBr

(CH3)2CCH2ClCl

(CH3)3COH

(CH3)3CBr(CH3)2C=CH2

(CH3)2C=CH2

(CH3)2C=CH2

(CH3)2C=CH2

(CH3)2C=CH2

(CH3)2C=CH2

H2PdH2OH2SO4

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.

CH2 CH

COOHCH2 C

H

COOHCH2 C

H

COOH


CH2CCOOH

HCH2C

COOH

HCH2C

H

COOH

13.68 Draw the polymer using the steps in Example 13.7.

CH2 C

CH3

COOCH3

CH2 CCH3

COOCH3

CH2 CCH3

COOCH3


CH2CCOOCH3

CH3

CH2CCOOCH3

CH3

CH2CCH3

COOCH3

13.69 Draw the polymers using the steps in Example 13.7.

CH2CCH2

H

CH2CCH2

H

CH2C

H

CH2a.

CH3 CH3 CH3

CH2 CCH2CH3

HCH2 C

CH2CH3

HCH2 C

CH2CH3

H


CH2CCl

CNCH2C

Cl

CNCH2C

CN

Clb. CH2 C

Cl

CNCH2 C

Cl

CNCH2 C

Cl

CN



Chapter 13–23

CH2CCl

ClCH2C

Cl

ClCH2C

Cl

Clc.

CH2 CCl

ClCH2 C

Cl

ClCH2 C

Cl

Cl


13.70 Draw the polymers using the steps in Example 13.7.

CH2 C

OCH3

HCH2 C

OCH3

HCH2 C

OCH3

H


CH2COCH3

HCH2C

OCH3

HCH2C

H

OCH3

a.

CH2 CCl

CO2CH3

CH2 CCl

CO2CH3

CH2 CCl

CO2CH3


CH2CCl

CO2CH3

CH2CCl

CO2CH3

CH2CCO2CH3

Cl

b.

CH2 CH

NHCOCH3

CH2 CH

NHCOCH3

CH2 CH

NHCOCH3


CH2CH

NHCOCH3

CH2CH

NHCOCH3

CH2C

NHCOCH3

H

c.

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

CH2 CBr

ClCH2 C

Br

ClCH2 C

Br

ClCH2 C

Br

Cl

Each one of these units is from the monomer:

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

CH2 CC

CH3CH2 CCH3

CCH2 C

CH3

CCH2 C

CH3

C

Each one of these units is from the monomer:

OO OOCH2CH3 OCH2CH3 OCH2CH3

OCH2CH3

O

13.73 Draw two resonance structures by moving the double bonds.

Cl Cl



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.



a. b.

chloro

p-chloroethylbenzene o-bromofluorobenzene

bromo

fluoro

ethyl



OH

propyl phenolm-propylphenol

a. H2N Br

aniline bromo

b.

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

a.

b.

c.

d.

Cl NO2

H2N NO2

Cl OH

Cl

butyl

2,5-dichlorophenol

ethyl

2,5-dichloro

nitro

chloro

nitro

NH2 on benzene ring = aniline


m-chloronitrobenzene

p-nitroaniline

o-butylethylbenzene


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.

a.

b.

c.

d.CH2CH3

CH3(CH2)3 (CH2)3CH3

CH3 I

2-bromo-3-iodotoluene

ethyl

ethyl


o-bromoethylbenzene

p-dibutylbenzenem-ethylaniline

2 butyl groups

Br bromo

H2N CH2CH3


Br bromoiodo

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.

NH2Cl

NH2

Cl

NH2

Cl

o-chloroaniline m-chloroaniline p-chloroaniline 13.80 Draw the structure of 2,4,6-trinitrotoluene (TNT).

CH3NO2

NO2

O2N

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

nitro

a.

NO2

CH2CH2CH3

CH2CH2CH2CH3

CH2CH2CH2CH3

b.

butyl

butyl

propyl

p-nitropropylbenzene

m-dibutylbenzene

chloro


OH on benzene ring = phenolOHI

CH3Br

Cl

NH2

ClI

c.

d.

e.

iodoiodo

chloro

bromoCH3 on benzene ring = toluene

1 2

34

1 2

34

56

o-iodophenol

2-bromo-4-chlorotoluene

2-chloro-6-iodoaniline



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

nitro

a.

NO2 NO2

NO2

d.

nitro

ethyl

m-ethylnitrobenzene 1,3,5-trinitrobenzene

CH2CH3 O2N

b.

F OH

e.


o-difluorobenzene 2,4-dibromophenolF

fluoroBr

Br bromo

c.

CH3

p-bromotoluene

bromoBr


13.83 Draw the products of each reaction.

CH3CH3

Cl

CH3CH3

NO2

CH3CH3

SO3H

CH3 CH3a.

b.

c.

CH3 CH3

CH3 CH3Cl2FeCl3

HNO3H2SO4

SO3

H2SO4

13.84 Determine the reagents for the reactions in the sequence.

FeCl3

Cl

H2SO4

Cl

H2SO4O2N

Cl

PdO2N SO3H

Cl

H2N SO3H

A B C

D

Cl2 HNO3 SO3

H2

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

HNO3H2SO4

Br Br Br

NO2

NO2

NO2

Br


Chapter 13–27

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

+ Cl2FeCl3

Cl

A

Br

Br

Br

Br

+ Cl2FeCl3

Br

C

Br Br

Cl

BrBr

Br+

Cl

+ Cl2FeCl3

B

Br Br

Br Br

Cl+

Br

BrCl

+

Br

BrCl

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

C CCH2

H H


H

CH3CH2CH2CH2CH2

H

CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2COOHCC

H

CH3CH2CH2CH2CH2

H

C CCH3CH2CH2CH2CH2CH2CH2CH2

H H

CH2CH2CH2CH2CH2CH2CH2COOH

a.

H2, Pd

Partial hydrogenation adds hydrogen to one of the double bonds.

+

CH2CH2CH2CH2CH2CH2CH2CH2CH2COOHCH3CH2CH2CH2CH2CH2CH2CH2

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



C CCH3CH2CH2CH2CH2CH2CH2CH2

H CH2CH2CH2CH2CH2CH2CH2COOH

Hc.

one possibility:

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

C C

(CH2)7COOH

H

C

H

CH

H

CCH

H

CH3(CH2)2CH2a.

C C

H

H

C

(CH2)7COOH

CH

H

CCH

H

CH3(CH2)2CH2b.

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.

CH2CH2CH2OH OCH3 OCH3HOa. b. c.

This compound could be an antioxidant because it has an OH

group bonded to the aromatic ring.

an ethernot an antioxidant

an alcoholnot an antioxidant

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

C CH H

(CH2)7COOHCH3(CH2)5C C

H (CH2)7COOH

HCH3(CH2)5C C

H H

(CH2)8COOHCH3(CH2)4a. b. c.

cispalmitoleic acid

transstereoisomer 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

ClCl

nonpolarthe dipoles cancel

polar

13.101

CH2CR

HCH2C

R

HCH2C

H

RR = (CH2)4CH3d. polymerization:

a. CH2=CH(CH2)4CH37 C chain1-heptene

b. CH2=CH(CH2)4CH3 CH3(CH2)5CH3H2

c. CH2=CH(CH2)4CH3 CH3CH(OH)CH2CH2CH2CH2CH3H2O

13.102

CH2CR

HCH2C

R

HCH2C

H

RR = (CH2)7CH3d. polymerization:

a. CH2=CH(CH2)7CH310 C chain1-decene

b. CH2=CH(CH2)7CH3 CH3(CH2)8CH3H2

c. CH2=CH(CH2)7CH3 CH3CH(OH)CH2CH2CH2CH2CH2CH2CH2CH3H2O

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).

trans-3-hexene

C CH

CH2CH2CH3

H

CH3C C

CH3CH2

CH2CH3

H

H

cis-2-hexene



13.104 Determine the two monomer units for Saran.

CH2 CCl

HCH2 C

H

ClCH2 C

Cl

ClCH2 C

H

ClCH2 C

Cl

Cl

A B A B

CH2 CCl

Cl

A B

+


Chapter 13 Unsaturated...

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