Download Chapter 14 Solutions

Chapter 14–1
Chapter 14 Organic Compounds That Contain Oxygen, Halogen, or Sulfur
Solutions to In-Chapter Problems
14.1
Label the –OH groups, –SH groups, halogens, and ether oxygens in each compound.
b. The OH on the benzene ring of salmeterol is part of a phenol, so it is not an alcohol.
14.2
To determine whether an alcohol is 1o, 2o, or 3o, locate the C with the OH group and count the
number of C’s bonded to it. A 1o alcohol has the OH group on a C bonded to one C, and so forth,
as in Example 14.1.
14.3
Use the definition in Example 14.1 and Answer 14.2 to label the hydroxyl groups.
14.4
Alcohols have stronger intermolecular forces and therefore higher boiling points than
hydrocarbons of comparable size and shape.
Organic Compounds with O, X, or S 14–2
14.5
Hydrocarbons are insoluble in water. Low molecular weight alcohols (< 6 C’s) are water soluble,
but higher molecular weight alcohols (≥ 6 C’s) are insoluble in water.
14.6
To name alcohols using the IUPAC system, follow the steps in Example 14.2:
[1] Find the longest carbon chain that contains the carbon bonded to the OH group.
[2] Number the carbon chain to give the OH group the lower number, and apply all other rules of
nomenclature.
14.7
Work backwards to draw the structure corresponding to each name.
Chapter 14–3
14.8
To draw the products of the dehydration of each alcohol, follow the steps in Example 14.3.
• First find the carbon bonded to the OH group, and then identify all carbons with H’s bonded
to this carbon.
• Remove the elements of H and OH from two adjacent C’s, and draw a double bond between
these C’s in the product.
• When two different alkenes are formed, the major product has more C’s bonded to the C=C.
14.9
The Zaitsev rule states that when two different alkenes are formed, the major product has more
C’s bonded to the C=C.
Organic Compounds with O, X, or S 14–4
14.10
B has no H on the carbon adjacent to the carbon with the OH group, so H2O cannot be lost.
14.11
Draw the products when each alcohol is oxidized, as in Example 14.4.
• RCH2OH (1o alcohols) are oxidized to RCHO, which are then oxidized to RCOOH.
• R2CHOH (2o alcohols) are oxidized to R2CO.
• R3COH (3o alcohols) are not oxidized because they have no H atom on the C with the OH.
14.12
Ethylene glycol is oxidized to a dicarboxylic acid.
14.13
Draw the three constitutional isomers.
Chapter 14–5
14.14
Look at the functional groups to determine the strength of the intermolecular forces, as in
Example 14.5. The stronger the forces, the higher the boiling point.
14.15
Low molecular weight ethers are water soluble. If the ether has more than five carbons, the ether
is insoluble in water.
14.16
Name each ether as in Example 14.6.
14.17
Work backwards from the IUPAC name to the structure.
Organic Compounds with O, X, or S 14–6
14.18
b. Desflurane is organic, and by “like dissolves like,” organic compounds are soluble in organic
solvents.
14.19
Draw the three-dimensional structure of halothane. Recall that solid lines represent bonds in the
plane of the page, wedges represent bonds in front, and dashed lines represent bonds behind.
14.20
Classify each alkyl halide as 1°, 2°, or 3°.
• A primary (1o) alkyl halide has a halogen on a carbon bonded to one carbon.
• A secondary (2o) alkyl halide has a halogen on a carbon bonded to two carbons.
• A tertiary (3o) alkyl halide has a halogen on a carbon bonded to three carbons.
14.21
Label the alkyl halide and each alcohol as 1°, 2°, or 3°.
Chapter 14–7
14.22
The boiling point of alkyl halides increases with the size of the alkyl group as well as the size of
the halogen.
14.23
Give the IUPAC name for each compound. Always start by finding the longest chain.
14.24
Work backwards from the name to the structure.
Organic Compounds with O, X, or S 14–8
14.25
Chlorofluorocarbons (CFCs) have the general molecular structure CFxCl4 – x.
Hydrochlorofluorocarbons (HCFCs) contain the elements of H, Cl, and F bonded to carbon,
whereas hydrofluorocarbons (HFCs) contain the elements of H and F bonded to carbon.
a. CF3Cl is a CFC.
b. CHFCl2 is an HCFC.
c. CH2F2 is an HFC.
14.26
Name the thiols using the steps in Example 14.7.
14.27
When thiols are oxidized, they form disulfides. Disulfides are converted back to thiols with a
reducing agent.
14.28
Solutions to End-of-Chapter Problems
14.29
To determine whether an alcohol is 1o, 2o, or 3o, locate the C with the OH group and count the
number of C’s bonded to it. A 1o alcohol has the OH group on a C bonded to one C, and so forth,
as in Example 14.1.
Chapter 14–9
14.30
To determine whether an alcohol is 1o, 2o, or 3o, locate the C with the OH group and count the
number of C’s bonded to it. A 1o alcohol has the OH group on a C bonded to one other C, and so
forth, as in Example 14.1.
OH
CH2CH2CH3
a.
CH3
b.
c.
(CH3)3CCH2CH2OH
CH3
CH3CH2CH2CHOH
d.
CH3CH2CH2CCH2OH
CH3
C bonded to 1C
1° alcohol
C bonded to 3 C's
3° alcohol
C bonded to 2 C's
2° alcohol
C bonded to 1 C
1° alcohol
14.31 Classify each alkyl halide as 1°, 2°, or 3°.
• A primary (1o) alkyl halide has a halogen on a carbon bonded to one carbon.
• A secondary (2o) alkyl halide has a halogen on a carbon bonded to two carbons.
• A tertiary (3o) alkyl halide has a halogen on a carbon bonded to three carbons.
14.32
Classify each alkyl halide as 1°, 2°, or 3°.
• A primary (1o) alkyl halide has a halogen on a carbon bonded to one carbon.
• A secondary (2o) alkyl halide has a halogen on a carbon bonded to two carbons.
• A tertiary (3o) alkyl halide has a halogen on a carbon bonded to three carbons.
C bonded to 1 C
1° alkyl halide
CH3
CH2Br
CH3C CHCH2CH2CCH CH2
Cl
C bonded to 3 C's
3° alkyl halide
14.33
Draw a structure that fits each description.
14.34
Draw a structure that fits each description.
OH
a.
CH3CH2CHCH2CH CH2
2° alcohol
CH2Cl
b.
O
cyclic ether
c.
CH3CH2CH
CH3
1° alkyl halide
Organic Compounds with O, X, or S 14–10
14.35
Draw the structure of the six constitutional isomers of molecular formula C5H12O that contain an
ether functional group.
14.36
Draw the structure of the four constitutional isomers of molecular formula C3H6Br2.
Br
Br
CH3 C CH3
CH3 CHCH2Br
Br
14.37
Br
CH3CH2C H
Br
BrCH2CH2CH2Br
Chapter 14–11
14.38
2 C's: ethyl group
5 C's: pentyl group
CH3CH3 O CH2CH2CH2CH2CH3
a.
b.
ether
2
CH3CH2CH2CH2CHCH3
6C's: hexanethiol
ethyl pentyl ether
2-hexanethiol
SH
thiol
14.39
14.40
a.
3
1
O
14.41
C bonded to 2 C's
b. 2° alcohol
c. 3-ethylcyclohexanol
H
To name alcohols using the IUPAC system, follow the steps in Example 14.2:
[1] Find the longest carbon chain that contains the carbon bonded to the OH group.
[2] Number the carbon chain to give the OH group the lower number, and apply all other rules of
nomenclature.
Organic Compounds with O, X, or S 14–12
14.42
To name alcohols using the IUPAC system, follow the steps in Example 14.2:
[1] Find the longest carbon chain that contains the carbon bonded to the OH group.
[2] Number the carbon chain to give the OH group the lower number, and apply all other rules of
nomenclature.
3
a.
CH3CH2CHCH2CH2CHCH3
OH
CH3
7 C's in the longest chain
heptanol
6
CH3CH2CHCH2CH2CHCH3
OH
OH group at C3
CH3
methyl group at C6
6-methyl-3-heptanol
Chapter 14–13
OH group at C1
1
CH2OH
b.
CH2OH
CH3CH2CH2CHCH2CH3
CH3CH2CH2CHCH2CH3
2-ethyl-1-pentanol
2
5 C's in the longest chain
pentanol
ethyl group at C2
methyl group at C3
c.
3
CH3(CH2)3CHCH3
CH3(CH2)3CHCH3
1
7 C's in the longest chain
heptanol
d.
OH group at C1
OH
OH
HOCH2CHCH2OH
HOCH2CHCH2OH
3 C's in the longest chain
propanol
OH
CH2CH3
e.
3-methyl-1-heptanol
CH2CH2OH
CH2CH2OH
1, 2,3-propanetriol
OH groups at C1, C2, C3
OH
CH2CH3
1-ethyl-2-methylcyclopentanol
OH and ethyl at C1
CH3
CH3
5 C's in the ring
cyclopentanol
methyl at C2
OH at C1
HO
f.
CH3
CH3
6 C's in the ring
cyclohexanol
14.43
HO 1
3 CH3
3,4-dimethylcyclohexanol
4 CH3
methyls at C3 and C4
Work backwards to draw the structure corresponding to each name.
Organic Compounds with O, X, or S 14–14
14.44
Work backwards to draw the structure corresponding to each name.
OH
a. 3-methyl-3-pentanol
CH3CH2CCH2CH3
5 C chain with OH at C3
and methyl at C3
b. 4-methyl-2-pentanol
2
c. 2,4-dimethyl-2-hexanol
4
OH
CH3
OH
4
CH3CCH2CHCH2CH3
CH3 CH3
6 C chain with OH at C2
and methyls at C2 and C4
e. 3,5-dimethylcyclohexanol
6 C in a ring with OH at C1
and methyls at C3 and C5
2
CH3CHCH2CHCH3
5 C chain with OH at C2
and methyl at C4
d. 1,3-propandiol
3 C chain with OH at C1
and C3
CH3
3
1
3
HOCH2CH2CH2OH
H3C 5
1 OH
3
CH3
OH
f. 6,6-diethyl-4-nonanol
9 C chain with OH at C4
and 2 ethyls at C6
CH2CH3
CH3CH2CH2CHCH2CCH2CH2CH3
4
6
CH2CH3
Chapter 14–15
14.45
Name each ether as in Example 14.6.
14.46
Name each ether as in Example 14.6.
1
a. CH3O CH2CH2CH3
3 C's in the longer chain
propane
CH3
3C's in ring
cyclopropane
14.47
1-methoxypropane
(or methyl propyl ether)
CH3
CH3O CHCH2CH3
2
4 C's in the longer chain
butane
OCH3
3
methoxy group
b. CH3O CHCH2CH3
c.
2
CH3O CH2CH2CH3
methoxy group at C2
OCH3
methoxycyclopropane
methoxy group
Draw the structures and then name the isomers.
2-methoxybutane
Organic Compounds with O, X, or S 14–16
14.48
Draw the structures and then name the isomers.
CH3
CH3CH2CH2CH2OH
1-butanol
methyl at C2
OH at C1
methyl at C2
OH
CH3CH2CHCH3
2-butanol
14.50
Name each compound using the IUPAC system.
14.51
2-methyl-2-propanol
OH at C2
Name each compound using the IUPAC system.
CH2CH3
10 C's in the longest chain
F at C3
ethyl at C5
5-ethyl-3-fluorodecane
CH3
OH
14.49
a. CH3CH2CHCH2CHCH2CH2CH2CH2CH3
OH at C1
CH3CCH3
OH at C2
F
2-methyl-1-propanol
CH3CHCH2OH
b.
SH
CH3CH2
c.
Cl
5 C's in the ring
Cl at C1
ethyl at C3
1-chloro-3-ethylcyclopentane
Work backwards from the name to draw each structure.
CH3CHCH3
3 C's in the longest chain
SH at C2
2-propanethiol
Chapter 14–17
14.52
Work backwards from the name to draw each structure.
a. dicyclohexyl ether
d. 3-methyl-1-pentanethiol
O
5 C chain
two 6 C rings
SH at C1
chlorine at C1
CH3
CH3C OCH2CH3
b. 2-ethoxy-2-methylpropane
e. 1-chloro-2-methylpropane
CH3
3 C chain
3 C chain
ethoxy at C2
ethoxy at C2
c. propyl iodide
3 C chain
14.53
CH3 methyl at C2
CH3CH2CHCH2CH2SH
f. 2-bromo-3-methylheptane
CH3CH2CH2 I
ClCH2CHCH3
CH3
CH3CHCHCH2CH2CH2CH3
Br CH3
7 C chain
iodine at C1
methyl at C2
bromine at C2
methyl at C3
Alcohols have higher boiling points than hydrocarbons of comparable size and shape. The
boiling points of alkyl halides increase with the size of the alkyl group and the size of the
halogen.
a. CH3CH2CH2I (larger halogen)
b. HOCH2CH2OH (two hydroxyl groups)
c. CH3CH2CH2OH (can hydrogen bond)
14.54
CH3CH2OH < CH3CH2CH2OH < CH3CH2CH2CH2OH
14.55
C has the highest boiling point because it has an OH group that can intermolecularly hydrogen
bond. B is a hydrocarbon with the weakest intermolecular forces, so it has the lowest boiling
point. In order of increasing boiling point:
CH3CH2CH2CH2CH3
B
14.56
<
CH3CH2OCH2CH3
A
<
CH3CH2CH2CH2OH
C
The boiling point of D is higher than the boiling point of F, even though F has a higher molecular
weight, because D is capable of hydrogen bonding to itself, whereas F is not. E is an alkane, so it
will have the lowest boiling point.
butane (E) < 1-chlorobutane (F) < 1-butanol (D)
14.57
Ethanol is a polar molecule capable of hydrogen bonding with itself and water. Dimethyl ether is
polar, but cannot hydrogen bond with itself. The stronger intermolecular forces make the boiling
point of ethanol higher. Both ethanol and dimethyl ether have only two carbons and can
hydrogen bond to H2O, so both are water soluble.
14.58
1,6-hexanediol is much more water soluble than 1-hexanol because 1,6-hexanediol has two OH
groups that can hydrogen bond to water, whereas 1-hexanol has only one OH group.
14.59
B (CH3CH2CH2CH2OH) has an OH group capable of intermolecular hydrogen bonding, so its
boiling point is higher than the boiling point of A (CH3CH2CH2CH2SH), which has weaker
intermolecular forces. Thus, the boiling point of A is 98 °C and the boiling point of B is 118 °C.
Organic Compounds with O, X, or S 14–18
14.60
X[CH3CH2CHOHCH2CH3] is 3-pentanol whereas Y [CH3CH2CH2CHSHCH3] is 2-pentanethiol.
X is water soluble because it has an OH group capable of hydrogen bonding to water. Y is unable
to hydrogen bond to water and is therefore insoluble in water.
14.61
To draw the products of dehydration of each alcohol, follow the steps in Example 14.3.
14.62
To draw the products of dehydration of each alcohol, follow the steps in Example 14.3.
a. (CH3)2CHCH2CH2CH2OH
H2SO4
(CH3)2CHCH2CH
CH2
This C loses a H to form the product.
CH2CH3
b.
CH3CH2CCH2CH3
H2SO4
H CH2CH3
CH3 C C CH2CH3
OH
This C loses a H to form the product.
OH
c. CH3CHCH2CH2CH2CH2CH2CH3
Either C can lose a H.
d.
CH3
H2SO4
OH
Either C can lose a H.
H2SO4
CH3CH CHCH2CH2CH2CH2CH3
2 C's bonded to the double bond
major product
CH3
3 C's bonded to
the double bond
major product
+
CH2
2 C's bonded to
the double bond
+
CH2 CH(CH2)5CH3
1 C bonded to the double bond
Chapter 14–19
14.63
Draw the products of dehydration, as in Example 14.3.
b. The carbons of both double bonds are bonded to an equal number of C’s; as a result, roughly
equal amounts of both isomers are formed.
14.64
The Zaitsev rule states that the major product in elimination is the alkene that has more alkyl
groups bonded to the double bond.
OH
a.
CH3
H2SO4
+
CH3
CH3
b. Both products are formed in equal amounts because the carbons of both double bonds are
bonded to an equal number of C’s.
14.65
Work backwards to determine what alcohol can be used to form each product.
14.66
Work backwards to determine what alcohol can be used to form each product.
CH3
OH
CH3
CH3
CH3
a.
b.
CH3
C C
CH3
CH3
CH3
H
CH3 C
OH
C CH3
CH3 CH3
14.67
Work backwards to determine what alcohols can be used to form propene. The OH group can be
bonded to either the middle or the end carbon.
14.68
Work backwards to determine what alcohols can be used to form 2-methylpropene. The OH
group can be bonded to either the middle or the end carbon.
CH3
CH3 C=CH2
CH3
CH3 C CH3
OH
CH3
or
CH3 C CH2OH
H
Organic Compounds with O, X, or S 14–20
14.69
Draw the products when each alcohol is oxidized, as in Example 14.4.
• RCH2OH (1o alcohols) are oxidized to RCHO, which are then oxidized to RCOOH.
• R2CHOH (2o alcohols) are oxidized to R2CO.
• R3COH (3o alcohols) are not oxidized because they have no H atom on the C with the OH.
14.70
Draw the products when each alcohol is oxidized, as in Example 14.4.
• RCH2OH (1o alcohols) are oxidized to RCHO, which are then oxidized to RCOOH.
• R2CHOH (2o alcohols) are oxidized to R2CO.
• R3COH (3o alcohols) are not oxidized because they have no H atom on the C with the OH.
OH
O
[O]
[O]
a.
CH3
CH3
2° alcohol
ketone
1° alcohol
14.71
CH3(CH2)8COH
carboxylic acid
Draw the product of oxidation.
O
COH
carboxylic acid
1° alcohol
O
[O]
b. CH3(CH2)8CH2OH
c.
CH2OH
[O]
d.
(CH3CH2)3COH
3° alcohol
No reaction
Chapter 14–21
14.72
Classify the OH groups and draw the products of oxidation.
O
1°
2°
CH2OH
H C OH
a.
2°
HO C H
b.
CH2OH
C OH
H C OH
H C OH
HO C H
[O]
HO C H
H C OH
2°
H C OH
H C OH
CH2OH
1°
CH2OH
C OH
O
O
CH2OH
C OH
H C OH
c.
C O
[O]
HO C H
O C
H C OH
C O
CH2OH
C OH
O
14.73
Work backwards to determine what alcohol can be used to prepare each carbonyl compound.
14.74
Work backwards to determine what alcohol can be used to prepare each carbonyl compound.
CH3
O
CH3
a.
ketone
OH
2° alcohol
b. CH3CH2CH2CH2CH2CO2H
CH3CH2CH2CH2CH2CH2OH
1° alcohol
carboxylic acid
14.75
Draw the products of each reaction.
14.76
Draw the products of each reaction.
CH3
OH
a. CH3CHCH2CHCH2CH2CH3
CH3
OH
b. CH3CHCH2CHCH2CH2CH3
H2SO4
K2Cr2O7
CH3
CH3
CH3CHCH CHCH2CH2CH3
CH3
O
CH3CHCH2CCH2CH2CH3
+
CH3CHCH2CH CHCH2CH3
Organic Compounds with O, X, or S 14–22
14.77
14.78
CH3 O
CH3 H
CH3 C
CH3 C C
C OH
CH3 CH3
CH3 CH3
2° alcohol
ketone
14.79
Draw the structure of the alcohol that fits the description. Since no reaction occurs with an
oxidizing agent, the compound must be a 3° alcohol. Since only one alkene is formed when
treated with sulfuric acid, all carbons adjacent to the C–OH must be identical.
14.80
(CH3CH2)3COH is a tertiary alcohol and will not react with an oxidizing agent such as K2Cr2O7,
whereas CH3(CH2)6OH is a primary alcohol and will undergo oxidation. The test tube that shows
a color change from red-orange to green is the one that contains the CH3(CH2)6OH.
14.81
Draw the disulfides formed by thiol oxidation.
14.82
Draw the disulfide formed by thiol oxidation.
COOH
H2N C H
[O]
COOH
COOH
H2N C CH2S SCH2 C NH2
H
CH2SH
H
14.83
Draw the thiols formed by disulfide reduction.
14.84
Draw the thiol formed by disulfide reduction.
CH3CH2
CH2CH3
S S
[H]
CH3CH2CHCH2CH2CHCH2CH3
SH
SH
Chapter 14–23
14.85
Write a balanced equation for the combustion of diethyl ether.
14.86
Write a balanced equation for the combustion of methanol.
2 CH3OH +
3 O2
2 CO2
+
4 H2O
methanol
14.87
a. CFCs are chlorofluorocarbons with a general formula of CFxCl4 – x; HCFCs have fluorine,
chlorine, and hydrogen bonded to carbon; and HFCs have only hydrogen and fluorine bonded
to carbon.
b. CFCs destroy the ozone layer whereas HCFCs and HFCs decompose more readily before
ascending to the ozone layer.
14.88
An example of a CFC is CF3Cl. The widespread use of CFCs has been detrimental to the
environment because they destroy the ozone layer that protects earth’s surface from ultraviolet
radiation.
14.89
PEG is capable of hydrogen bonding with water, so PEG is water soluble. PVC cannot hydrogen
bond to water, so PVC is water insoluble, even though it has many polar bonds.
14.90
F
H
a.
F C F
F C O C
H
H
F
C F
F
b. Halogenated ethers are now used in place of diethyl ether as anesthetics because they are less
flammable than diethyl ether and do not cause nausea in patients.
14.91
The greater the blood alcohol level, the greater the color change from red-orange to green with
the breathalyzer.
14.92
2,3-Butanediol is needed to prepare 2,3-butanedione by an oxidation reaction.
OH OH
CH3 C C CH3
H H
2,3-butanediol
[O]
O O
CH3C CCH3
2,3-butanedione
Organic Compounds with O, X, or S 14–24
14.93
Draw the product of the oxidation of propylene glycol.
14.94
Draw the product of the oxidation of lactic acid.
OH
O
[O]
CH3 C COOH
CH3C COOH
H
lactic acid
14.95
Draw the oxidation products formed from ethanol. Antabuse blocks the conversion of
acetaldehyde to acetic acid, and the accumulation of acetaldehyde makes people ill.
14.96
In order to make straight hair curly, the disulfide bonds in the protein chains are reduced to free
SH groups. The hair is turned around curlers and then oxidized to reform new disulfide bonds
that help the hair to maintain its new curls.
14.97
Ether molecules have no H’s on O’s capable of hydrogen bonding to other ether molecules, but
the hydrogens in water can hydrogen bond to the oxygen of the ether.
14.98
Figure a shows the oxygen atom of the alcohol hydrogen bonding with a hydrogen atom from
water. Figure b shows the hydrogen atom of the alcohol hydrogen bonding to the oxygen atom of
the water.
H
a.
O
CH3CH2
ethanol
H
O
H water
b.
O
CH3CH2
ethanol
H
O H
H
water
Chapter 14–25
14.99
Answer each question about the alcohol.
14.100 Answer each question about the alcohol.
a. 2-methyl-1-heptanol
b. 1° alcohol
CH3
c.
CH3
d.
e.
CH3CH2CH2CH2CH2CHCH2OH
CH3
H2SO4
CH3CH2CH2CH2CH2CHCH2OH
CH3CH2CH2CH2CH2C CH2
CH3
K2Cr2O7
CH3CH2CH2CH2CH2CHCOOH
CH3CH2CH2CH2CH2CH2CH2CH2OH
constitutional isomer
f.
CH3CH2CH2CH2CH2CH2CH2OCH3
constitutional isomer
14.101
14.102
OH H
C C
H H
H2SO4
+
C C
H
H
H
C C
H