To synthesize macromolecules with complex architecture one can use various approaches: apply differe — Physical Chemistry — Kinetics Chemistry Question
Co-polymerization
To synthesize macromolecules with complex architecture one can use various approaches: apply different types of polymerization, vary initiators, solvents and reaction conditions, copolymerize different monomers, as well as modify the obtained polymers. Some examples of copolymers are given in the table hereunder:
Type of copolymer | Schematic structure | Abbreviation
Block | AAAAAAAAAAAABBBBBBBBBBBBBB | poly(A)-block-poly(B)
Alternating | ABABABABABABABAB | poly(A-alt-B), poly(AB)
Statistical | AABABAABBBAABBBABAABABAAB | poly(A-stat-B)
Graft | AAAAAAAAAAAAAAAAAAAAAAAAAAAA (with B branches on the side) | poly(A)-graft-poly(B)
Gradient | AAAAABAAABAABBABABBBBABBBB | poly(A-grad-B)
While developing copolymerization technique it is important to take into account relative reactivity of monomers. Kinetics of copolymerization can be described by a set of elementary reactions with corresponding rate constants. In the case of binary radical copolymerization four elementary reactions of chain propagation should be considered (end-unit model):
R1 + M1 → R1 (k11)
R2 + M1 → R1 (k21)
R1 + M2 → R2 (k12)
R2 + M2 → R2 (k22)
Relative reactivity of monomers in copolymerization is characterized by the ratio of the rate constants of their addition to a given macroradical: r1 = k11 / k12, and r2 = k22 / k21. These ratios are referred to as copolymerization constants (r value is always between zero and unity). For instance, for styrene and maleic acid anhydride the copolymerization constants are 0.04 and 0.01, respectively. Sometimes, the same approach is applied to define constants of binary ionic copolymerization.
Complete equations of polymerization reactions below and draw structures of compounds X1 – X7. Give both detailed and short formulas of all copolymers. In short formulas represent styrene units as St, ethylene oxide units as EO, vinyl alcohol units as VA, and maleic anhydride units as MA. Use abbreviations from the above table when necessary.
[VISUAL]
Model Answer
a)
Reaction sequence:
1. Styrene + Na (metallic sodium/sodium naphthalenide) → X1 (disodium polystyrene dianion, initiating styrene polymerization from both ends)
2. X1 + Styrene → X2 (propagating polystyrene dianion)
3. X2 + Ethylene oxide → X3 (addition of ethylene oxide blocks to both ends of the polystyrene block)
4. X3 + termination agent → X4: poly(EO)-block-poly(St)-block-poly(EO)
b)
Reaction sequence:
1. Poly(vinyl alcohol) + a Na (where a << n) → alcoholate intermediate
2. Intermediate + Styrene → grafting of styrene chains from the oxygen anions along the PVA backbone → X5: poly(VA)-graft-poly(St)
c)
Reaction sequence:
1. Styrene + Maleic Anhydride in the presence of AIBN (initiator) at temperature T → Alternating copolymerization occurs → X6: poly(St-alt-MA)
2. X6 + NaOH/H2O → Hydrolysis of the maleic anhydride units to maleate units → X7: poly(St-alt-Ma) (where Ma is maleate)
Calculate the average length of a chain of units A in the polymer obtained by radical copolymerization of equimolar mixture of two monomers of the same reactivity.
Model Answer
Monomers possess equal reactivity (r1 = r2 = 1). Thus, fraction of units A in the polymer is the same as that of monomers in the reaction mixture and is equal to ½. Besides, distribution of units along the chain is random. So we conclude that fractions of dyads AA, AB, BA and BB are equal (1/4).
Solution 1.
Let us consider a long polymeric chain of N units. It contains N/2 of units A (with accuracy to one unit). The total number of dyads AB and BA is (N–1)/2, as there are N–1 dyads in the whole chain. The number of blocks in the chain exceeds the total number of dyads AB and BA by 1, and is equal to (N+1)/2, half of the blocks being composed of A. Thus, there are (N+1)/4 blocks of A in the chain. Then the average number of A units per block is: ((N+1)/2) : ((N+1)/4) ≈ (N/2) : (N/4) = 2.
Solution 2.
Average lengths of blocks composed of A and B are equal due to symmetry of problem with respect to permutation (A, B). In the chain containing N units there are (N+1)/2 ≈ N/2 blocks. Thus, the average length of block is N:(N/2) = 2.