chapter 13 unsaturated...
TRANSCRIPT
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
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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
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.
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
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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
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Unsaturated Hydrocarbons 13–4
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
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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
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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.
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.
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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
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Unsaturated Hydrocarbons 13–8
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
CH2CH2CH2CH2CH2CH2CH2COOHCC
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
Join these 2 C's. Join these 2 C's.
CH
CH
HCH
CH
HCH
CH
H
Cl Cl Cl
CH2 CH
Cl
c. CH2 CH
Cl
CH2 CH
Cl
Join these 2 C's. Join these 2 C's.
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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.
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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
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
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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°
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Unsaturated Hydrocarbons 13–12
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
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.
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
7 C's in the longest chainheptene
1 2
3 4 5 6 7double bond at C1
CH2 C CH2CH3
CH2CH2CH2CH2CH3
ethyl at C2
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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
2 methyls at C5 and C6
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
double bonds at C1 and C5
6 C's in the longest chainhexadiene
6 5 43
2 methyls at C3
CH2 CHCH2 CCH3
CH3
CH=CH2
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.
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
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Unsaturated Hydrocarbons 13–14
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
4 methyls at C2 and C5
(CH3)3CC CC(CH3)3
CH3CC CCCH3
CH3
CH3 CH3
CH3
CH3CC CCCH3
CH3
CH3 CH3
CH352
e.
Answer: 3-butyl-2-heptene
7 C's in the longest chainheptene
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
double bonds at C2 and C7
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
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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
double bonds at C1 and C4
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
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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.
5 carbon ring double bond at C1
methyls at C1 and C2a. 1,2-dimethylcyclopentene1
2
CH3
CH3
1
2
CH3
CH3
8 carbon chain triple bond at C2
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
7 carbon chain triple bond at C2
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
10 carbon chain triple bond at C1
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
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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
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Unsaturated Hydrocarbons 13–18
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.
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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
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Unsaturated Hydrocarbons 13–20
13.56 Determine if the molecules are constitutional isomers, stereoisomers or identical.
CH3anda.
same molecular formulasame connectivity
identical
same molecular formulasame connectivity
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
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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
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Unsaturated Hydrocarbons 13–22
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
Join these 2 C's. Join these 2 C's.
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
Join these 2 C's. Join these 2 C's.
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
Join these 2 C's. Join these 2 C's.
CH2CCl
CNCH2C
Cl
CNCH2C
CN
Clb. CH2 C
Cl
CNCH2 C
Cl
CNCH2 C
Cl
CN
Join these 2 C's. Join these 2 C's.
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Chapter 13–23
CH2CCl
ClCH2C
Cl
ClCH2C
Cl
Clc.
CH2 CCl
ClCH2 C
Cl
ClCH2 C
Cl
Cl
Join these 2 C's. Join these 2 C's.
13.70 Draw the polymers using the steps in Example 13.7.
CH2 C
OCH3
HCH2 C
OCH3
HCH2 C
OCH3
H
Join these 2 C's. Join these 2 C's.
CH2COCH3
HCH2C
OCH3
HCH2C
H
OCH3
a.
CH2 CCl
CO2CH3
CH2 CCl
CO2CH3
CH2 CCl
CO2CH3
Join these 2 C's. Join these 2 C's.
CH2CCl
CO2CH3
CH2CCl
CO2CH3
CH2CCO2CH3
Cl
b.
CH2 CH
NHCOCH3
CH2 CH
NHCOCH3
CH2 CH
NHCOCH3
Join these 2 C's. Join these 2 C's.
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
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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.
chloro
p-chloroethylbenzene o-bromofluorobenzene
bromo
fluoro
ethyl
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.
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
OH on benzene ring = phenol
m-chloronitrobenzene
p-nitroaniline
o-butylethylbenzene
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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
CH3 on benzene ring = toluene
o-bromoethylbenzene
p-dibutylbenzenem-ethylaniline
2 butyl groups
Br bromo
H2N CH2CH3
NH2 on benzene ring = aniline
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
NH2 on benzene ring = aniline
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
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Unsaturated Hydrocarbons 13–26
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.
OH on benzene ring = phenol
o-difluorobenzene 2,4-dibromophenolF
fluoroBr
Br bromo
c.
CH3
p-bromotoluene
bromoBr
CH3 on benzene ring = toluene
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
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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
CH2CH2CH2CH2CH2CH2CH2COOHCC
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:
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Unsaturated Hydrocarbons 13–28
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.
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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
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Unsaturated Hydrocarbons 13–30
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
+
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