Download CHAPTER 12: ANSWERS TO SELECTED PROBLEMS

CHAPTER 12: ANSWERS TO SELECTED PROBLEMS
SAMPLE PROBLEMS (“Try it yourself”)
12.1
CH3–CH2–CH2–O–CH2–CH2–CH3
12.2
CH3–CH2–CH2–CH2–O–CH2–CH2–CH2–CH3
O
12.3
CH3 CH2
C
O
O
12.4
CH3 CH2 CH2 CH2
C
NH2
O
12.5
O
P
O
CH3
O
12.6
HO–CH2–CH2–CH2–O–CH2–CH2–CH2–O–CH2–CH2–CH2–OH
O
NH2
12.7
H
N
CH3
O
CH
C
H
N
C
NH2
CH2
O
CH
C
H
HO
C
CH3
O
CH
C
OH
H
CH3
12.8
N
(CH2) 4
O
O
C
CH3
O
C
CH3
O
O
C
OH
CH3
12.9
CH3
12.10
and
OH
CH3
C
HO
CH2 CH2 CH
CH3
O
CH2
C
OH
and
HO
CH3
CH3
O
12.11 NH3
and
HO
CH3
12.12
CH3
C
C
CH2 CH2 CH2 CH3
O
CH2
C
and
O
HO
CH3
CH3
O
12.13
CH3 CH2 NH3
and
O
C
CH3
O
12.14
CH3
C
O K
and
HO
CH3
SECTION PROBLEMS
Section 12.1
12.1
CH3
a) CH3–CH2–O–CH2–CH2–CH2–CH3
c) CH3 CH2
b) CH3–CH2–CH2–O–CH2–CH2–CH3
(Note that usually there are several ways to draw the product of
a condensation reaction, depending on how you orient the reactants.
Your answers may look somewhat different from mine, but still be correct.)
12.2
a) ethyl pentyl ether
b) dimethyl ether
O
CH
CH2
CH2
CH3
Section 12.2
O
O
12.3
a) CH3
C
b) CH3 CH2
O
C
CH3
O
CH
CH3
CH3
CH3 O
12.4
a) CH3 CH2
N
C
b)
H
O
CH2
C
N
CH2
CH3
12.5 Molecule “a” will give a pH above 7, but molecule “b” will not. Molecule “b” is an
amide, and amides are not bases. Molecule “a” is an amine, and amines are bases.
O
O
a) CH3 CH2 CH2 CH2
O
O
O
O
12.6
P
P
b) CH3
O
O
C
CH3
CH3
Section 12.3
12.7
HO
O
O
12.8
HO
CH
CH3
C
O
O
O
CH
CH3
C
OH
O
O
CH
CH3
C
OH
12.9
CH3
CH3
CH3
CH2
CH2
CH2
CH3
CH
O
CH3
CH
O
CH3
CH
O
NH2
CH
C
NH
CH
C
NH
CH
C
OH
12.10 To make the product, first arrange the molecules so that they alternate:
HO CH2 CH2 OH
HO
O
O
C
C
HO CH2 CH2 OH
OH
HO
Then carry out the condensation reactions to form the final product:
HO CH2 CH2
O
O
O
C
C
O
CH2 CH2
O
O
O
C
C
Section 12.4
CH3
12.11 a) CH3
and
OH
HO
CH2 CH
b) Two molecules of
CH3
OH
O
12.12 a) CH3 CH2 CH2
C
and
OH
HO
CH3
O
b)
and
OH
HO
C
H
CH3 O
12.13 a) CH3
C
C
OH
and
NH3
CH3
O
CH3
b) CH3
N
H
and
HO
C
CH2 CH2 CH2 CH2 CH3
OH
O
O
C
C
OH
O
12.14 a) CH3 CH2 OH
and
HO
P
O
O
O
b) CH3 CH2
C
O
and HO P
OH
O
O
O
12.15 a) CH3
C
and
OH
O
CH3
b) CH3
CH2 CH3
HS
CH
SH
CH3
CH3
CH
CH
and
C
HO
CH2 CH3
Section 12.5
12.16
CH3
O
CH2
C
O
and
HO
CH3
O
12.17
C
O
and
NH3 CH2 CH3
12.18 A saponification reaction is the reaction of an ester with a strong base (NaOH or KOH),
to form an alcohol and the sodium or potassium salt of a carboxylic acid.
12.19 The answer to this problem is virtually the same as the answer to Problem 12.16. The
only difference is that you make the sodium salt of the carboxylate ion. The products are:
CH3
CH3
CH3
CH
CH
O
CH2
C
O
Na
and
HO
CH3
12.20 A soap is an ionic compound that contains a long-chain carboxylate ion and a sodium or
potassium ion.
CUMULATIVE PROBLEMS (Odd-numbered problems only)
12.21 A condensation reaction is a reaction in which two molecules combine to form a larger
molecule. In the process, one molecule loses H and the other loses OH, and the H and OH
combine to form water.
12.23 A dehydration removes H and OH from one molecule, converting a C–C single bond into
a double bond. A condensation removes H from one molecule and OH from a different
molecule, and links the remaining pieces into a single organic compound.
OH
12.25 a) CH3 CH
CH3
CH2
CH3
b) CH OH
+
HO
CH3
CH
CH3
+
CH3
CH3
CH
CH
CH3
CH3
H2O
CH3
O
CH
+
H2O
CH3
12.27 In a dehydration, an alcohol loses the OH group and a hydrogen atom from the carbon
adjacent to the functional group carbon. Methanol does not contain a carbon atom adjacent to
the functional group carbon, so it cannot be dehydrated.
CH3 CH2
OH
12.29
CH3 CH2 OH
O
CH3 CH2 CH
+
CH2 CH3
CH3 CH2 CH
CH2 CH3
+ H2O
12.31 Only the organic product is shown here. Water is also a product of any condensation
reaction.
O
a) CH3 CH2 CH2
C
O
CH2 CH2 CH3
O
b)
C
O
CH3 CH
CH2 CH3
c) The structures of the reactants are:
O
CH3 CH2 CH2 CH2 CH2
C
OH
OH
and
CH3
CH
CH3
The structure of the product is:
O
CH3 CH2 CH2 CH2 CH2
C
CH3
O
CH
CH3
O
12.33
CH3
O
C
O
CH2 CH2
C
O
CH3
O
12.35
CH2
O
C
(CH2) 14CH3
O
CH
O
C
(CH2) 14CH3
O
CH2
O
C
(CH2) 14CH3
12.37 As before, only the organic product is shown here. Water is also a product of these
reactions.
O
a) CH3 CH2 CH2
CH3
b) CH
CH3
C
NH2
O
NH
C
CH
CH2 CH3
CH3
c) The structures of the reactants are:
CH3 CH2 NH2
and
HO
The structure of the product is:
O
CH3 CH2 NH
C
O
H
C
H
O
12.139 a) CH3
CH2
O
CH2 CH3
C
OH
+
H
N
CH3 CH2
CH2 CH3
C
CH2 CH3
O
H
+
N
CH2 CH3
H
O
b)
CH3 CH2
C
CH2 CH3
OH
H
+
N
CH2 CH3
CH3 CH2
O
CH2 CH3
C
N
CH2 CH3
+ H2O
12.41 In order for an amine to condense with a carboxylic acid, the amine must have at least
one hydrogen atom that is directly bonded to the nitrogen atom. Triethylamine does not have a
hydrogen atom bonded to nitrogen (the nitrogen is bonded to three carbon atoms), so it cannot
undergo condensation reactions.
O
12.43
NH2
O
C
CH2 CH2
C
NH2
12.45 Note that acetic acid reacts with the alcohol group in salicylic acid, not the carboxylic
acid group.
O
C
O
carboxylic acid group
OH
C
O
HO
OH
O
C
CH3
O
phenol group
O
O
O
P
b) CH3 CH2 CH2
O
O
O
P
O
O
O
CH2 CH2 CH2
O
C
CH3
CH3
O
12.49
P
O
O
12.47 a) CH3 CH2 CH2
OH
O
P
O
O
C
CH3
O
12.51
CH3
O
P
O
CH3
O
CH3
CH3
12.53 HO CH CH2 CH
O
12.55
HO CH2
CH3
O
CH
CH3
CH2 CH
CH
12.57 HO CH2 CH2
12.61
NH2
O
CH
CH3
CH2 CH
OH
CH3
CH2
OH
O
12.59
CH3
CH2
O
CH
C
C
NH
NH2 CH2 CH2 NH
O
O
CH2 CH2
CH2
O
CH
C
O
O
C
C
O
C
NH
O
CH2 CH2
CH2
O
CH
C
C
O
O
CH2 CH2
C
OH
OH
NH CH2 CH2 NH
O
O
C
C
OH
12.63 There are three different ways that we can arrange two molecules of glycine and one
molecule of valine. We can put the valine on the left side, in the middle, or on the right side:
Option #1: valine – glycine – glycine
Option #2: glycine – valine – glycine
Option #3: glycine – glycine – valine
Here is option #1:
CH3
CH3
CH
O
NH2
CH
C
O
NH CH2
C
O
NH CH2
C
OH
Here is option #2:
CH3
NH2 CH2
O
CH3
CH
O
C
NH
CH
C
O
NH CH2
C
OH
Here is option #3:
CH3
O
NH2 CH2
C
NH CH2
O
CH3
CH
O
C
NH
CH
C
OH
12.65 You must draw the two molecules of acetic acid with their carboxylic acid groups facing
one another.
O
CH3
C
O
OH
+
C
HO
O
CH3
CH3
O
C
O
CH3
C
+ H2O
an anhydride
Note that this type of reaction does NOT normally occur.
12.67 a)
CH3 CH2
O
CH2 CH2 CH3
CH3
b) CH3
CH
CH3 CH
H2O
CH3 CH2
O
O
CH3
c)
+
H
+ H2O
CH3 O
C
CH3
CH3 CH2
C
CH
CH3
O
CH3 + H2O
O
+
OH
CH2 CH
C
CH3 CH2 NH
C
H
+
OH
HO CH3
O
N
+ H2O
CH3 CH2
C
OH
+
C
H
N
O
O
e)
HO
CH3 O
CH3 CH
O
d)
HO CH2 CH2 CH3
+
CH3
C
CH2 CH
OH
NH CH3
+
H2O
CH3 CH2 NH2
+
HO
C
CH2 CH3
O
O
f) CH3 CH2 CH2
O
P
O
CH3 CH2 CH2 OH
+ H2O
+
HO
O
O
12.69 a) CH3 CH2 OH
HO CH2 CH2 CH3
+
CH3
b) CH3 CH
P
O
OH
CH3
+
O
C
H
CH3 O
c) CH3 CH
CH2 CH
C
O
+
HO CH3
O
d) CH3 CH2
C
O
H
+
H
N
O
e) CH3 CH2 NH3
+
O
C
CH2 CH3
12.71 Only the amide group on the right side of the asparagine structure can be hydrolyzed.
The products are a carboxylic acid and ammonia.
O
C
NH2
CH
O
OH
O
CH2
C
C
NH2
the amide group
+
H2O
NH2
CH
OH
CH2
O
C
OH
+
NH3
the products (a carboxylic acid and ammonia)
O
O
12.73 a) CH3
O
C
CH2 CH3
S
O
b)
+
H2O
CH3
C
OH
O
C
O
+
CH2 CH3
HS
O
P
O
+ H2O
O
C
OH
+
HO
O
P
O
O
12.75 The organic reactant contains two ester groups, both of which can be hydrolyzed.
O
CH3
O
C
O
O
CH2
C
ester group
O CH3
+
2 H2O
CH3 OH
+
HO
C
O
CH2
C
OH
+
HO CH3
ester group
The first and third products are actually the same molecule (methanol), so you can also write this
reaction as:
O
CH3
O
C
O
CH2
O
C
O CH3
+
2 H2O
O
12.77
2 CH3 OH
+
O
C
2 CH3 OH
+
HO
O
CH2
C
C
OH
O
CH2
C
(Both carboxylic acid groups
are ionized at pH 7.)
O
12.79 There are two amide functional groups in the reactant. Both of them will be hydrolyzed.
O
CH3
C
NH2
CH3
O
CH
C
NH
OH
CH2
O
CH
C
amide group
NH
CH
CH3
CH2
O
CH
C
OH
amide group
Here are the structures of the three amino acids that will be formed, drawn in their unionized
forms.
O
CH
3
C
NH2
CH3
O
CH
C
OH
NH2
OH
CH2
O
CH
C
CH
OH
NH2
CH3
CH2
O
CH
C
OH
12.81 At pH 7, all of the carboxylic acid and amine groups will be ionized.
O
CH3
C
NH3
CH3
O
CH
C
O
NH3
O
CH2
O
CH
C
O
12.83 a) CH3 CH2
O
CH
O
NH3
CH3
CH2
O
CH
C
O
P
+
OH
P
HO
O
O
O
O
(Note: you can also draw the first product as CH3 CH2
O
P
O
O
b) CH3 CH2 OH
+
HO
O
P
O
P
O
CH2
O
O
O
C
12.85
OH
OH
12.87 a) CH3–CH2–O–CH2–CH3
b) CH3–CH2–O–CH2–CH3
c) CH3–CH2–O–CH2–O–CH2–CH3
12.89 There are eleven functional groups in this molecule!
CH3
alcohol
HO
amine
CH3
N
CH3
OH
alcohol
alkene
C
OH
OH
phenol
O
ketone
alkene
OH
alcohol
O
ketone
NH2
O
amide
O
)
O
12.91 Look at the functional group. Carboxylic acids, phenol, and thiols are acidic and produce
acidic solutions when they dissolve in water. Amines are basic and produce basic solutions
when they dissolve in water. The other functional groups you have studied are neither acidic nor
basic, so they do not affect pH.
a) basic: this is an amine
b) neutral: this is an amide
c) neutral: this is an ester
d) acidic: this is a carboxylic acid
e) neutral: this is an alcohol
f) neutral: this is an aldehyde
g) basic: this is an amine
h) acidic: this is a phenol
12.93 For the condensation reaction, the balanced equation (using molecular formulas) is:
2 C2H6O → C4H10O + H2O
When you use 10.0 g of ethanol, you form 1.96 g of water.
For the dehydration reaction, the balanced equation is:
C2H6O → C2H4 + H2O
When you use 10.0 g of ethanol, you form 3.91 g of water. (Use the formula weights of ethanol
and water as a conversion factor, and don’t forget to multiply the formula weight of ethanol by 2
in the condensation reaction.)
12.95
CH2
CH2
+
H2O
CH3 CH2 OH
compound A
O
CH3 CH2 OH
oxidation
CH3
C
H
compound A
compound B
O
O
CH3
C
H
oxidation
compound B
CH3
C
OH
compound C
O
O
CH3
C
OH
compound C
+
HO CH2 CH3
compound A
condensation
CH3
C
O
CH2 CH3
compound D
+
H2O