Lodge and Hiemenz, Polymer Chemistry, 3rd Edition, Solutions Manual
1.1. The ratio of B:I is 2:1; therefore, roughly 10 B’s to 5 I’s. To illustrate %’s of B, show 11
B’s: 5 cis, 5 trans, and 1 vinyl. The 5 I’s may all be shown as cis. B’s and I’s are in blocks;
the isomers of B are random.
1.2. Consider 100 repeat units:
47.2 + 44.9 = 92.1 units with 4 backbone C’s and 7.9 units with 2 backbone C’s:
92.1 units ´ 4 C’s/unit = 368 C’s
7.9 units ´ 2 C’s/unit = 16 C’s
∴
(368 + 16) C's / 100 units = 48.6
C atoms
7.9 ethyl groups/100 units ethyl
1.3.
1.4. (a) polyethylene
(b) polyoxymethylene
(c) poly(oxy-1,4-phenylene)
(d) poly(vinyl butyral)
1
, Lodge and Hiemenz, Polymer Chemistry, 3rd Edition, Solutions Manual
(e) poly(vinyl acetate)
(f) poly[(1-methoxycarbonyl)ethylene]
1.5. (a) R = cyclohexanone tetrapropionic acid for b = 4 and bicyclohexanone octapropionic
acid for b = 8.
b = number of arms in star
y = degree of polymerization of each linear chain
(b) 1 – L = number of equivalents of reacted groups per mole monomer
Q + L = total number of equivalents of “ends” per mole monomer
number of reacted groups
∴ ratio = =y
number of ends
(c) For a molecule with b arms
1− L
M n = bM n,arm + M central and M n,arm = yM o = M
Q+ L o
M central
Mb =
b
⎡ 1− L ⎤
∴ Mn = b ⎢ Mo + Mb ⎥
⎣ Q+ L ⎦
The repeat unit has 6 C’s, 1 O’s, 1 N, and 11 H so M0 = 113.
The central unit for b = 4 has the formula C18H26O9, so Mcentral = 386 and Mb = 96.5.
The central unit for b = 8 has the formula C36H50O18, so Mcentral = 770, and Mb = 96.
(d) Mn = 2449 for b = 4 and 7462 for b = 8.
1.6. By dividing mi by Mi, we obtain the number of moles for each fraction, ni. We then utilize
the equations
∑n M i i ∑n Mi i
2
∑n Mi
3
i
Mn = i
, Mw = i
and M z = i
∑n i ∑n Mi i ∑n Mi i
2
i i i
to obtain Mn = 29.1 kDa, Mw = 46.1 kDa and Mz = 62.5 kDa.
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, Lodge and Hiemenz, Polymer Chemistry, 3rd Edition, Solutions Manual
1.7. We consider the equation for the viscosity average molecular weight
⎛ ∑ ni M i1+a ⎞
1a
Mν = ⎜ i ⎟ .
⎜ ∑ ni M i ⎟
⎝ i ⎠
We obtain Mv = 662 kDa for acetone, Mv = 665 kDa for chloroform and Mn = 597 kDa, Mw =
671 kDa. So Mn < Mv,acetone < Mv,chloroform < Mw. If we consider the meaning of a, it quantifies
the dependence of the intrinsic viscosity on molecular weight. The higher value means that the
volume occupied by a chain, and consequently the solution viscosity, is more sensitive to the
larger polymers in the distribution.
1.8. Example solution (inspired by the characters of Marvel Comics):
Character Mass (kg) wi wiMi
Jean Gray 52 0.11 5.7
Rogue 54 0.11 6.2
Bruce Banner 58 0.12 7.1
Elektra 59 0.12 7.3
Spider-Man 75 0.16 11.8
Magneto 86 0.18 15.7
Wolverine 88 0.19 16.5
For the listed characters,
Mn = ∑
Mi
= 67.5 kg; M w = ∑ wi M i = 70.6 kg
7
Mw
D= = 1.04
Mn
Based on Ð, the selected distribution is quite narrow.
1.9. We’ll present two approaches to solve this problem. First,
z z+1 miz ⎛ zm ⎞
wi = exp ⎜ − i ⎟ and M w = ∑ wi mi .
(
Γ z + 1 Mn
z+1
)
⎝ Mn ⎠ i
To obtain an analytical expression for Mw that does not include a summation, we convert the
sum to an integral according to the rule
k' k'
∑ = ∫ dn :
n=k k
∞
z z+1 1
∞
⎛ zm ⎞
so M w = ∫ w( m) m dm = z+1 ∫
m z+1 exp ⎜ − ⎟ dm .
0 (
Γ z + 1 Mn 0 ) ⎝ Mn ⎠
With the aid of an integral table, we find that
∞
( ) (
Γ n +1 ).
∫x exp −ax dx =
n
n+1
0 a
Applying this to our expression for Mw, we obtain
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, Lodge and Hiemenz, Polymer Chemistry, 3rd Edition, Solutions Manual
Mw =
z z+1 1 Γ z+2 ( )
( )
Γ z + 1 M nz+1 z M z+2
n ( )
If z only assumes integer values, then G(z+1) = z! and G(z+2) = (z+1)! = (z+1)z!. Then the
expression for Mw simplifies to
z +1
Mw = Mn
z
Another option is to consider moments of distributions, as presented in the book:
∞
µ k = ∑ xi i k = ∫ i k P ( i )di .
i 0
Mw is proportional to the ratio of the second to the first moment of the distribution
∞
z z+1 m z ⎛ zm ⎞ ∞
µ1 = ∫ exp ⎜ − ⎟ dm = ∫ M n w m dm = M n . ( )
(
0 Γ z + 1 Mn
z
)
⎝ Mn ⎠ 0
Substituting this into the moment ratio for Mw, we get:
∞
z z+1 m z−1 ⎛ zm ⎞
∫ m Γ ( z + 1) M exp ⎜ −
2
z ⎟ dm ∞
⎛ zm ⎞
µ2 ⎝ Mn ⎠ z z+1 1
z+1 ∫
0
Mw = = n
= m z+1 exp ⎜ − ⎟ dm .
µ1 Mn ( )
Γ z + 1 Mn 0 ⎝ Mn ⎠
1.10. Find the maximum of wi w.r.t. Mi:
dwi z z+1 1 ⎛ − zM / M z − zM / M ⎞
= z+1 ⎜
zM i z−1e i n − Mize i n ⎟ = 0
dM i Γ(z + 1) M n ⎝ Mn ⎠
Dividing through all the common terms we reach
⎛ z−1 1 ⎞
⎜ Mi − M Mi ⎟ = 0
z
⎝ n ⎠
and therefore at the maximum of wi
Mi = Mn
1.11. The half-width at half-height is about 3,500, so s ≈ 3000. From eqn 1.7.16
1/2
⎛M ⎞
= 48,916 (1.008 − 1)
1/2
σ = M n ⎜ w − 1⎟ = 4,400
⎝ Mn ⎠
This is in the ballpark, but not exact.
1.12. (a) poly(methacrylic acid) (T) or poly[1-(carboxy)-1-methylethylene] (I)
A chain growth process (free radical) through the carbon-carbon double bond; initiation and
termination steps are not specified.
(b) poly(tetramethylene adipamide) (T) or poly(iminoadipoyliminobutane-1,4-diyl) (I)
4
, Lodge and Hiemenz, Polymer Chemistry, 3rd Edition, Solutions Manual
A step growth (polycondensation) process between a diamine and a diacid chloride.
(c) poly(6-hydroxyhexanoic acid) (T) or poly(oxycarbonyl pentamethylene) (I)
A step-growth (self-condensation) polymerization of an a-hydroxy, w-carboxy alkane.
(d) poly(acrylonitrile) (T) or poly(1-cyanoethylene) (I)
A chain growth process (free radical) through the carbon-carbon double bond; initiation and
termination steps are not specified.
(e) poly(tetramethylene phenylurethane) (T) or poly(oxytetramethyleneoxycarbonylimino-
1,4-phenyleneiminocarbonyl) (I)
A step-growth polymerization (strictly, not a condensation) between a diisocyanate and a diol.
Note that in reality the NH groups can also react with the NCO groups to produce some
crosslinking.
(f) poly(alanine) (T) or poly[imino(1-methyl-2-oxo-ethylene)] (I)
A step-growth (self-condensation) polymerization of an amino acid.
1.13. The molecular ion will include one silver ion (108 g/mol), leaving 1098 for the polymer.
As Mo = 104, it is tempting to infer a degree of polymerization of 10, leaving 58 for the mass
of the endgroups. This corresponds exactly to one butyl group and one hydrogen atom,
suggesting an anionic polymerization. Imagine this polystyrene in the all trans backbone
conformation. We are interested in the configuration of the side groups. There are two
possibilities for each repeat unit, leading to 210 = 1024 possible structures. Note that the two
ends of the chain are different, so we do not need to worry about overcounting.
1.14. We can write the molecular structure as
5
1.1. The ratio of B:I is 2:1; therefore, roughly 10 B’s to 5 I’s. To illustrate %’s of B, show 11
B’s: 5 cis, 5 trans, and 1 vinyl. The 5 I’s may all be shown as cis. B’s and I’s are in blocks;
the isomers of B are random.
1.2. Consider 100 repeat units:
47.2 + 44.9 = 92.1 units with 4 backbone C’s and 7.9 units with 2 backbone C’s:
92.1 units ´ 4 C’s/unit = 368 C’s
7.9 units ´ 2 C’s/unit = 16 C’s
∴
(368 + 16) C's / 100 units = 48.6
C atoms
7.9 ethyl groups/100 units ethyl
1.3.
1.4. (a) polyethylene
(b) polyoxymethylene
(c) poly(oxy-1,4-phenylene)
(d) poly(vinyl butyral)
1
, Lodge and Hiemenz, Polymer Chemistry, 3rd Edition, Solutions Manual
(e) poly(vinyl acetate)
(f) poly[(1-methoxycarbonyl)ethylene]
1.5. (a) R = cyclohexanone tetrapropionic acid for b = 4 and bicyclohexanone octapropionic
acid for b = 8.
b = number of arms in star
y = degree of polymerization of each linear chain
(b) 1 – L = number of equivalents of reacted groups per mole monomer
Q + L = total number of equivalents of “ends” per mole monomer
number of reacted groups
∴ ratio = =y
number of ends
(c) For a molecule with b arms
1− L
M n = bM n,arm + M central and M n,arm = yM o = M
Q+ L o
M central
Mb =
b
⎡ 1− L ⎤
∴ Mn = b ⎢ Mo + Mb ⎥
⎣ Q+ L ⎦
The repeat unit has 6 C’s, 1 O’s, 1 N, and 11 H so M0 = 113.
The central unit for b = 4 has the formula C18H26O9, so Mcentral = 386 and Mb = 96.5.
The central unit for b = 8 has the formula C36H50O18, so Mcentral = 770, and Mb = 96.
(d) Mn = 2449 for b = 4 and 7462 for b = 8.
1.6. By dividing mi by Mi, we obtain the number of moles for each fraction, ni. We then utilize
the equations
∑n M i i ∑n Mi i
2
∑n Mi
3
i
Mn = i
, Mw = i
and M z = i
∑n i ∑n Mi i ∑n Mi i
2
i i i
to obtain Mn = 29.1 kDa, Mw = 46.1 kDa and Mz = 62.5 kDa.
2
, Lodge and Hiemenz, Polymer Chemistry, 3rd Edition, Solutions Manual
1.7. We consider the equation for the viscosity average molecular weight
⎛ ∑ ni M i1+a ⎞
1a
Mν = ⎜ i ⎟ .
⎜ ∑ ni M i ⎟
⎝ i ⎠
We obtain Mv = 662 kDa for acetone, Mv = 665 kDa for chloroform and Mn = 597 kDa, Mw =
671 kDa. So Mn < Mv,acetone < Mv,chloroform < Mw. If we consider the meaning of a, it quantifies
the dependence of the intrinsic viscosity on molecular weight. The higher value means that the
volume occupied by a chain, and consequently the solution viscosity, is more sensitive to the
larger polymers in the distribution.
1.8. Example solution (inspired by the characters of Marvel Comics):
Character Mass (kg) wi wiMi
Jean Gray 52 0.11 5.7
Rogue 54 0.11 6.2
Bruce Banner 58 0.12 7.1
Elektra 59 0.12 7.3
Spider-Man 75 0.16 11.8
Magneto 86 0.18 15.7
Wolverine 88 0.19 16.5
For the listed characters,
Mn = ∑
Mi
= 67.5 kg; M w = ∑ wi M i = 70.6 kg
7
Mw
D= = 1.04
Mn
Based on Ð, the selected distribution is quite narrow.
1.9. We’ll present two approaches to solve this problem. First,
z z+1 miz ⎛ zm ⎞
wi = exp ⎜ − i ⎟ and M w = ∑ wi mi .
(
Γ z + 1 Mn
z+1
)
⎝ Mn ⎠ i
To obtain an analytical expression for Mw that does not include a summation, we convert the
sum to an integral according to the rule
k' k'
∑ = ∫ dn :
n=k k
∞
z z+1 1
∞
⎛ zm ⎞
so M w = ∫ w( m) m dm = z+1 ∫
m z+1 exp ⎜ − ⎟ dm .
0 (
Γ z + 1 Mn 0 ) ⎝ Mn ⎠
With the aid of an integral table, we find that
∞
( ) (
Γ n +1 ).
∫x exp −ax dx =
n
n+1
0 a
Applying this to our expression for Mw, we obtain
3
, Lodge and Hiemenz, Polymer Chemistry, 3rd Edition, Solutions Manual
Mw =
z z+1 1 Γ z+2 ( )
( )
Γ z + 1 M nz+1 z M z+2
n ( )
If z only assumes integer values, then G(z+1) = z! and G(z+2) = (z+1)! = (z+1)z!. Then the
expression for Mw simplifies to
z +1
Mw = Mn
z
Another option is to consider moments of distributions, as presented in the book:
∞
µ k = ∑ xi i k = ∫ i k P ( i )di .
i 0
Mw is proportional to the ratio of the second to the first moment of the distribution
∞
z z+1 m z ⎛ zm ⎞ ∞
µ1 = ∫ exp ⎜ − ⎟ dm = ∫ M n w m dm = M n . ( )
(
0 Γ z + 1 Mn
z
)
⎝ Mn ⎠ 0
Substituting this into the moment ratio for Mw, we get:
∞
z z+1 m z−1 ⎛ zm ⎞
∫ m Γ ( z + 1) M exp ⎜ −
2
z ⎟ dm ∞
⎛ zm ⎞
µ2 ⎝ Mn ⎠ z z+1 1
z+1 ∫
0
Mw = = n
= m z+1 exp ⎜ − ⎟ dm .
µ1 Mn ( )
Γ z + 1 Mn 0 ⎝ Mn ⎠
1.10. Find the maximum of wi w.r.t. Mi:
dwi z z+1 1 ⎛ − zM / M z − zM / M ⎞
= z+1 ⎜
zM i z−1e i n − Mize i n ⎟ = 0
dM i Γ(z + 1) M n ⎝ Mn ⎠
Dividing through all the common terms we reach
⎛ z−1 1 ⎞
⎜ Mi − M Mi ⎟ = 0
z
⎝ n ⎠
and therefore at the maximum of wi
Mi = Mn
1.11. The half-width at half-height is about 3,500, so s ≈ 3000. From eqn 1.7.16
1/2
⎛M ⎞
= 48,916 (1.008 − 1)
1/2
σ = M n ⎜ w − 1⎟ = 4,400
⎝ Mn ⎠
This is in the ballpark, but not exact.
1.12. (a) poly(methacrylic acid) (T) or poly[1-(carboxy)-1-methylethylene] (I)
A chain growth process (free radical) through the carbon-carbon double bond; initiation and
termination steps are not specified.
(b) poly(tetramethylene adipamide) (T) or poly(iminoadipoyliminobutane-1,4-diyl) (I)
4
, Lodge and Hiemenz, Polymer Chemistry, 3rd Edition, Solutions Manual
A step growth (polycondensation) process between a diamine and a diacid chloride.
(c) poly(6-hydroxyhexanoic acid) (T) or poly(oxycarbonyl pentamethylene) (I)
A step-growth (self-condensation) polymerization of an a-hydroxy, w-carboxy alkane.
(d) poly(acrylonitrile) (T) or poly(1-cyanoethylene) (I)
A chain growth process (free radical) through the carbon-carbon double bond; initiation and
termination steps are not specified.
(e) poly(tetramethylene phenylurethane) (T) or poly(oxytetramethyleneoxycarbonylimino-
1,4-phenyleneiminocarbonyl) (I)
A step-growth polymerization (strictly, not a condensation) between a diisocyanate and a diol.
Note that in reality the NH groups can also react with the NCO groups to produce some
crosslinking.
(f) poly(alanine) (T) or poly[imino(1-methyl-2-oxo-ethylene)] (I)
A step-growth (self-condensation) polymerization of an amino acid.
1.13. The molecular ion will include one silver ion (108 g/mol), leaving 1098 for the polymer.
As Mo = 104, it is tempting to infer a degree of polymerization of 10, leaving 58 for the mass
of the endgroups. This corresponds exactly to one butyl group and one hydrogen atom,
suggesting an anionic polymerization. Imagine this polystyrene in the all trans backbone
conformation. We are interested in the configuration of the side groups. There are two
possibilities for each repeat unit, leading to 210 = 1024 possible structures. Note that the two
ends of the chain are different, so we do not need to worry about overcounting.
1.14. We can write the molecular structure as
5
