1. The sodium naphthalene polymerization of methyl methacrylate is carried out in Benzene
and tetrahydrofuran solutions. Which solution will yield the highest polymerization rate?
Discuss the effect of solvent on the relative concentrations of the different types of
propagating centers.
Solution:
The polymerization of methyl methacrylate in the presence of sodium naphthalene is an
example of anionic polymerization. The stability of formation of anion is the key step in this
case. The anion formation is greatly enhanced by polar solvent such as tetrahydrofuran
(THF) as opposed to Benzene non-polar solvent.
The order is tetrahydrofuran (THF) is higher solvating power in solvents than Benzene
makes it more polar solvent. Increase in solvating power increases concentration of free Ions
propagating centers relative to the ion pairs, and increases the concentration of solvent-
separated ion pairs relative to contact ion pairs. The effect is relatively minor compared with
the effect of polymerization rate by changing concentrations of monomer and/or initiator.
Polymerization rate and polymer molecular weight increase with increasing solvent polarity
because there is a shift in concentrations from the unreactive (dormant) covalent species
toward the ion pair and free ions. Ionic polymerizations in general, commonly involve two
types of propagating species; an ion pair and free ion coexisting in.
Thus, the rate of polymerization in this case is faster in tetrahydrofuran (THF) solvent than in
Benzene
, 2. Assume that 1.0 103 mol of sodium naphthalene is dissolved in tetrahydrofuran and
then 2.0 mol of styrene is introduced into the system by a rapid injection technique. The
final total volume of the solution is 1 liter. Assume that the injection of styrene results in
instantaneous homogeneous mixing. It is found that half of the monomer is polymerized
in 2000 s. Calculate the propagation rate constant. Calculate the degree of
polymerization at 2000 and at 4000 s of reaction time.
Solution:
Assume first order kinetics;
d M
Rp k p M
dt
M d M t
M 0
M
k p dt
0
M 0
ln k pt
M
M 0
To find half-life period ( t1/2 ), substitute t = t1/2 and M
2
ln(2) k p t1/2
ln(2)
kp
t1/2
0.69314
kp
2000
k p 3.4657 104 Sec.-1
The propagation rate constant k p 3.4657 104 Sec.-1
and tetrahydrofuran solutions. Which solution will yield the highest polymerization rate?
Discuss the effect of solvent on the relative concentrations of the different types of
propagating centers.
Solution:
The polymerization of methyl methacrylate in the presence of sodium naphthalene is an
example of anionic polymerization. The stability of formation of anion is the key step in this
case. The anion formation is greatly enhanced by polar solvent such as tetrahydrofuran
(THF) as opposed to Benzene non-polar solvent.
The order is tetrahydrofuran (THF) is higher solvating power in solvents than Benzene
makes it more polar solvent. Increase in solvating power increases concentration of free Ions
propagating centers relative to the ion pairs, and increases the concentration of solvent-
separated ion pairs relative to contact ion pairs. The effect is relatively minor compared with
the effect of polymerization rate by changing concentrations of monomer and/or initiator.
Polymerization rate and polymer molecular weight increase with increasing solvent polarity
because there is a shift in concentrations from the unreactive (dormant) covalent species
toward the ion pair and free ions. Ionic polymerizations in general, commonly involve two
types of propagating species; an ion pair and free ion coexisting in.
Thus, the rate of polymerization in this case is faster in tetrahydrofuran (THF) solvent than in
Benzene
, 2. Assume that 1.0 103 mol of sodium naphthalene is dissolved in tetrahydrofuran and
then 2.0 mol of styrene is introduced into the system by a rapid injection technique. The
final total volume of the solution is 1 liter. Assume that the injection of styrene results in
instantaneous homogeneous mixing. It is found that half of the monomer is polymerized
in 2000 s. Calculate the propagation rate constant. Calculate the degree of
polymerization at 2000 and at 4000 s of reaction time.
Solution:
Assume first order kinetics;
d M
Rp k p M
dt
M d M t
M 0
M
k p dt
0
M 0
ln k pt
M
M 0
To find half-life period ( t1/2 ), substitute t = t1/2 and M
2
ln(2) k p t1/2
ln(2)
kp
t1/2
0.69314
kp
2000
k p 3.4657 104 Sec.-1
The propagation rate constant k p 3.4657 104 Sec.-1
