Chemistry, Student Solutions Manual
Chapter 17
Chapter 17 Macromolecules
Solutions to Problems in Chapter 17
17.1 Convert line drawings to structural formulas following the standard rules, adding a
C atom at each vertex and line end and adding –H bonds until each C atom has four
bonds. Identify functional groups from memory or from Table 17-1 in your
textbook:
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17.3 Identify functional groups from memory and convert structures to line drawings
following the standard rules:
17.5 Convert line and ball-and-stick drawings to structural formulas following the
standard rules, adding a C atom at each vertex and line end and adding –H bonds
until each C atom has four bonds. Identify linkage groups from memory or Table
17-1 in your textbook:
17.7 To draw examples of compound types, first identify the chemical formulas of its
functional groups. Remember that carbon always forms four bonds. There are many
possible correct answers to this problem. We provide just one example for each.
(a) An amine contains N bonded to C. One example is the tertiary amine
diethylmethylamine:
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(b) An ester contains C–CO2–C. There must be five additional C atoms, arranged in
any fashion:
(c) The functional group in an aldehyde is O=C–H, M = 29 g/mol. To have a molar
mass greater than 80, an aldehyde must contain at least four additional C atoms:
(d) The ether linkage is –O–, and phenyl is the benzene ring:
17.9 In a condensation reaction, two monomers react to form a larger unit, eliminating a
small molecule such as H2O in the process.
(a) The amide linkage in the centre of this molecule forms in a condensation
reaction between a carboxylic acid and an amine:
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(b) The ester linkage in this molecule forms in a condensation reaction between a
carboxylic acid and an alcohol:
(c) The ether linkage in this molecule forms in a condensation reaction between two
alcohols:
17.11 All amino acids can condense with one another in two ways. In addition, the extra
carboxylic acid of aspartic acid could undergo condensation, so there are three
possible products:
17.13 Polymers made from substituted ethylenes form by breakage of the C=C double
bond to form two new single C–C bonds to other monomers. In a copolymer, each
monomer occurs randomly:
17.15 To deconstruct an alkane polymer into its monomers, break C–C single bonds and
form double bonds:
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17.17 Polybutadiene forms from butadiene by a combination of double-bond breakage
and migration. The resulting polymer differs from polyethylene in having one C=C
double bond in each repeat unit, whereas polyethylene is entirely CH2 units
connected by C–C single bonds (each arrow represents shifting one electron):
17.19 To determine the monomers from which a condensation polymer has formed,
decompose the polymer into its monomeric parts and add components of a small
molecule, for example –H and –OH. Here, the structure shows a single component
whose repeat unit is (CH2)10, and the linkage is an amide, which breaks down into
an amine and a carboxylic acid:
17.21 Construct polyethylene oxide by breaking a C–O bond and linking the fragments:
17.23 Cross-linking leads to increased rigidity, because chemical bonds between polymer
chains restrict the ability of polymer chains to slide past one another. Thus,
relatively rigid tires have more extensive cross-linking than flexible surgeon’s
gloves.
17.25 The categories of polymers and their characteristic properties are as follows:
plastics, which exist as blocks or sheets; fibres, which can be drawn into long
threads; and elastomers, which can be stretched without breaking.
(a) Balloons must stretch, so they are made of elastomers.
(b) Rope is made of fibres.
(c) Camera cases are rigid, so they are made of plastics.
17.27 Dioctylphthalate is an example of a liquid plasticizer, which reduces the amount of
cross-linking in a polymer as well as adding a fluid component; both these changes
result in improved flexibility of the polymer.
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17.29 To draw the structures of sugars that are related to glucose, start with the glucose
structure and make modifications as needed. For α-talose, move the OH group on
carbons 2 and 4 from down positions to up positions.
17.31 To switch between α and β isomers, move the OH group on carbon 1 from a down
position to an up position (or vice versa when going from β to α). To draw βtalose, start with the structure of α-talose from Problem 17.29 and move the OH on
carbon 1 from a down position to an up position.
17.33 As described in your textbook, polymers of α-glucose coil upon themselves.
Glycogen is this type of polymer. Polymers of β-glucose, of which cellulose is an
example, form planar sheets. It is easier to see the distinction between the two
polymers by looking at them from a side-on view (as shown). In the drawings, the
dark solid lines indicate the direction of the continuing chain:
17.35 The complementary strands of DNA form from hydrogen-bond linkages between
specific nucleic acids. A pairs with T, and G pairs with C, so the complementary
sequence of A-A-T-G-C-A-C-T-G is T-T-A-C-G-T-G-A-C.
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17.37 The structure of DNA consists of a phosphate–sugar–nucleotide-base trio bonded
to others through its phosphate and sugar groups. Apart from the structure of the
nucleotide base, each unit is identical:
17.39 The strands of DNA fit together as a result of the hydrogen-bonding interactions
illustrated in Figure 17-27. The complementary sequence for A-T-C is T-A-G. The
backbones of complementary strands run in opposite directions. For clarity in
showing the hydrogen-bonding interactions, we represent the backbone with a
solid line rather than showing its details:
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17.41 Consult Figure 17-31 for the structures of the different amino acids, all of which
have the same amino acid backbone.
17.43 Hydrophilic side chains are characterized by the presence of N or O atoms that
generate polar bonds and hydrogen-bonding capability, or an S–H bond that is
polar. Among the structures shown in Problem 17.41, Tyr (O–H bond) and Glu
(CO2H group) are hydrophilic, whereas Phe and Met are hydrophobic.
17.45 To identify an amino acid, examine the side chain attached to the carbon atom
between the N atom and the C=O bond in the amino acid backbone: (a) The side
chain is just –H (hydrophobic), making this glycine, the simplest amino acid. (b)
The side chain is –CH2OH (hydrophilic side chain), so this is serine. (c) The side
chain is –CH2SH (hydrophilic side chain), so this is cysteine.
17.47 When two amino acids condense, the amino end of either molecule can link to the
carboxylic acid end of the other. A water molecule is eliminated, creating a C–N
bond. Three amino acids can combine in six different ways: A–B, B–A, A–C, C–
A, B–C, and C–B.
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17.49 The repeat unit in polyethylene is the ethylene molecule, C2H4, M = 28.1g/mol, so
a polymer with 744 repeat units has
M  (744)(28.1 g/mol)  2.09  104 g/mol.
17.51 There are four possible choices for each base in a DNA strand, so the number of
ways to connect 12 bases is 412 = 16 777 216 (this is an exact number, because 4 is
an exact number).
17.53 (a) The monomers from which hair spray is made are substituted ethylenes, which
polymerize in a free radical process in which each ethylene has an equal probability
of adding to the growing chain. Thus, this polymer will have a random
arrangement.
(b) The backbone of a polyethylene is a carbon chain with the substituents
connected to every other carbon atom. Here is a line structure of six monomer units:
(c) The backbone of this polymer is hydrophobic, but its side chains contain highly
polar C=O groups that interact with one another and with polar and hydrogenbonding groups on hair.
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17.55 A nucleotide is a combination of one base, one sugar, and one phosphate group. A
duplex is a pair of nucleotides bound together by hydrogen-bonding interactions.
The second nucleotide in a guanine duplex is cytosine.
17.57 The three steps of free radical polymerization are (1) initiation by a free radical
initiator, which generates a new free radical; (2) propagation, in which monomer
units add to the free radical end of the polymer chain; and (3) termination, in which
two free radical chains link together. The functional group on an ethylene
monomer does not participate in any of these processes:
17.59 All sugar molecules have multiple –OH functional groups, each of which can
participate in hydrogen-bond formation with water molecules. α-Glucose has five –
OH groups; in addition, its ring O atom can participate in formation of hydrogen
bonds.
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17.61 Nylons and proteins share one common feature: they form by condensation
reactions between carboxylic acids and amines, so their linkages are amide groups.
Otherwise, they are quite different. Nylons contain one or at most two different
monomers, each of which typically contains several carbon atoms that form part of
the backbone of the polymer. Proteins contain backbones that are absolutely
regular repetitions of amide–C–amide bonding, but the carbon atoms in the
backbone have a variety of substituent groups attached to them, generating the
immense variety of different proteins (compared with only a few different nylons).
17.63 Consult your textbook for the structures of the polymers, which indicate the
monomers from which they are made: (a) Kevlar is made from terephthalic acid
and phenylenediamine. (b) PET is made from ethylene glycol and terephthalic
acid. (c) Styrofoam is the common name for polystyrene, so it is made from
styrene.
17.65 All polyethylenes have the same empirical formula, (CH2)n, but whereas highdensity polyethylene has all straight chains that nest together readily, low-density
polyethylene has many side chains that cannot nest easily. Thus, low-density
polyethylene has more open space, accounting for the lower amount of CH2 groups
per unit volume. See Figure 17-9 for a visual representation.
17.67 The Watson–Crick model of DNA requires that there be a complementary base for
each base in a given strand: one G for every C, one C for every G, one A for every
T, and one T for every A. This requires that the molar ratios of A to T and G to C
be 1.0. Chargoff’s observations indicate that this relationship holds even though
DNA from different sources has different sequences, some A–T rich and others G–
C rich. These differences give rise to differences in the relative amounts of A, T vs.
G, C, but pairing always results in 1:1 mole ratios of A to T and G to C.
17.69 The bases of the mRNA molecule must be complementary to the bases of the
template DNA on which they are modelled, meaning A generates U, C generates
G, G generates C, and T generates A (remember that in RNA, U appears rather
than T). Thus, the sequence generated by this strand is the same as that of Strand B
except that U replaces T:
1 = cytosine, 2 = uracil, 3 = adenine, and 4 = guanine.
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17.71 All proteins contain both hydrophobic and hydrophilic amino acids. When a
protein is in contact with a hydrophilic medium such as aqueous solution, its
hydrophilic amino acids are most stable when facing outward, in contact with the
solvent, whereas its hydrophobic amino acids are most stable when facing inward,
in contact with the protein backbone. The opposite is the case when a protein is
immersed in a hydrophobic medium such as a cell wall. The tertiary structure of a
protein is the manner in which the individual amino acids orient themselves, so
solvent interactions are the primary determinants of this tertiary structure.
17.73 To draw the structure of a condensation product, start with the structures of the two
molecules undergoing condensation and then connect them together, eliminating a
small molecule such as water:
17.75 Polystyrene has the polyethylene backbone of carbon atoms, with a benzene ring
attached to every second carbon atom. A divinylbenzene monomer reacts with the
growing chain by means of one of its C=C double bonds, leaving the second C=C
double bond available to cross-link by becoming incorporated into another
polystyrene chain. Thus, the benzene rings form “bridges” between polystyrene
chains:
17.77 A cyclic product forms when one end of a monomer molecule can condense with
the other end of the same molecule. The monomer illustrated in this problem
contains five carbon atoms in addition to its terminal functional groups, so it can
easily form a ring containing six carbon atoms and one oxygen atom:
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17.79 To determine which amino acids are used to make a dipeptide, examine the
structure of the dipeptide and see what R groups it contains. The amine end of
aspartame contains an extra –CH2COOH group, characteristic of aspartic acid; the
carboxylic acid end of the molecule contains a benzene side group, characteristic
of phenylalanine.
17.81 The structure of lactic acid shows that it is bifunctional, with a carboxylic acid and
an alcohol group. These groups can condense to form an ester linkage, so the lactic
acid polymer is a polyester:
17.83 When amino acids condense to form polypeptides, each added amino acid reacts to
eliminate a water molecule. The chemical formula of alanine is C3H7NO2, and the
net reaction to form poly-Ala is n C3H7NO2 → n H2O + (C3H5NO)n.
Thus, the empirical formula of this polymer is C3H5NO, M = 71 g/mol.
Divide the molar mass of the polypeptide by the molar mass of the empirical
formula unit to obtain the number of repeat units:
1.20 × 103 g/mol
 17 repeat units of alanine.
71 g/mol
17.85 The monomers from which this alkyd forms are an alcohol with three –OH groups
and a carboxylic acid with two –COOH groups. These can condense together,
eliminating water molecules to form ester linkages. The third –OH allows the
possibility of branching in addition to linear structures, and the 2:3 stoichiometry
of alcohol to carboxylic acid indicates that all the functional groups undergo
condensation. Here is a small part of the structure, built around one starting alcohol
molecule:
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17.87 Polyethylene is a carbon backbone (with two C–H bonds to each C atom). The
butadiene–styrene polymer, while also containing an all-carbon backbone, has
benzene rings attached at intervals of approximately six carbon atoms, as well as a
double bond in the backbone at intervals of approximately six carbon atoms. The
extra bulk of the benzene rings and the geometrical changes where the double
bonds exist (trigonal planar as opposed to tetrahedral) make it impossible for the
carbon chains to stack as closely together in butadiene–styrene as in polyethylene.
With less close stacking, the dispersion forces holding the chains together are
weaker, so butadiene–styrene is not as rigid as polyethylene.
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