Polymer Chemistry I: Monomers, Chain-Growth and Step-Growth Polymerization, Metallocene and Controlled Methods
2. Free-radical addition polymerization and its steady-state kinetics
Chain-growth proceeds in three stages. Initiation: an initiator I (a peroxide or azo compound such as AIBN) decomposes, I → 2R•, rate constant k_d, and R• adds to a monomer. Only a fraction f (efficiency, about 0.3–0.8) of radicals starts chains, so the rate of initiation is Ri = 2f·k_d[I]. Propagation: M• + M → M•, rate constant k_p, repeated 10³–10⁵ times in about a second per chain, Rp = k_p[M][M•]. Termination: two radicals react by combination (coupling, giving one chain from two) or disproportionation (H transfer, giving two chains), with rate Rt = 2k_t[M•]² under the convention that k_t is defined per radical lost.
Steady-state hypothesis: radical concentration is tiny (10⁻⁸ to 10⁻⁷ mol/L) and constant, so Ri = Rt: 2f·k_d[I] = 2k_t[M•]², giving [M•] = (f·k_d[I]/k_t)^½. Then
Rp = k_p[M]·(f·k_d[I]/k_t)^½
(the long-chain approximation: monomer is consumed only in propagation)
Rp is first order in monomer and half order in initiator; doubling [I] raises Rp by only √2, the half-order being the signature of bimolecular termination. The overall activation energy is E = E_p + ½(E_d − E_t); since E_d is large (about 120–150 kJ/mol), rate rises steeply with temperature while Xn falls.
3. Kinetic chain length, chain transfer and the Mayo equation
The kinetic chain length ν is the average number of monomer units added per radical generated: ν = Rp/Ri = Rp/Rt = k_p[M]/(2(f·k_d·k_t[I])^½). With no transfer, Xn = 2ν if termination is entirely by combination (each chain carries two initiator fragments) and Xn = ν if by disproportionation. Numerical sense: Rp = 1.0×10⁻⁴ mol L⁻¹ s⁻¹ and Ri = 2.0×10⁻⁷ mol L⁻¹ s⁻¹ give ν = 500, so Xn = 1000 (combination) or 500 (disproportionation).
Chain transfer (to monomer, solvent S, initiator, polymer or a deliberate transfer agent) kills one chain but passes the radical on, so the rate is unchanged while chains shorten. The Mayo equation is 1/Xn = 1/Xn,0 + C_M + C_S[S]/[M] + C_I[I]/[M], where Xn,0 is the value without transfer and each C = k_tr/k_p. A Mayo plot of 1/Xn against [S]/[M] is a straight line of slope C_S. Mercaptans (C_S large) are used as molecular-weight regulators. Transfer to polymer makes branches (LDPE short- and long-chain branching).
Ionic and coordination polymerization (outline). Cationic: electron-donating substituents stabilise the carbocation (isobutylene, vinyl ethers) with Lewis acid plus co-initiator (BF₃/H₂O), at low temperature, termination by transfer. Anionic: electron-withdrawing groups stabilise the carbanion (styrene, acrylonitrile, dienes) with initiators such as n-butyllithium; in a pure system there is no termination (living), so Xn = [M]₀/[I]₀ at full conversion, chains stay active, and block copolymers follow by sequential monomer addition. Coordination (Ziegler–Natta): heterogeneous TiCl₃/TiCl₄ with an aluminium alkyl gives stereoregular isotactic polypropylene and linear polyethylene by insertion of the monomer at the transition-metal–carbon bond (Cossee–Arlman picture).
4. Step-growth polymerization and the Carothers equation
In step-growth, monomers first form dimers, trimers and so on; any two species can react, and high polymer appears only at very high conversion. Let p be the extent of reaction (fraction of the functional groups, of the limiting type, that have reacted). For an equimolar AA + BB or an A–B system, with N₀ initial molecules and N present at p, Xn = N₀/N, and since N = N₀(1 − p):
Xn = 1/(1 − p)
(Carothers equation, equimolar). p = 0.90 → Xn = 10; p = 0.98 → 50; p = 0.99 → 100; p = 0.999 → 1000.
For a polyester with mean structural-unit mass M₀ = 113 g/mol reacted to p = 0.99, Xn = 100 and Mn = 100 × 113 = 11,300 g/mol (ignoring end groups). Reaching Mn of useful engineering polymers therefore needs p above 0.99, which in turn demands purity, no side reactions and removal of the by-product (water, or methanol in ester interchange) to drive the equilibrium.
Stoichiometric imbalance. With r = N_A/N_B ≤ 1 (ratio of the numbers of A and B groups, A being deficient), Xn = (1 + r)/(1 + r − 2rp). At p → 1, Xn = (1 + r)/(1 − r): for r = 0.98, Xn = 1.98/0.02 = 99, so a 2% excess of one monomer caps the chain length near 100 however long the reaction runs. A deliberate trace of monofunctional reagent likewise controls Mn and stabilises it against later chain growth.
Network formation. For monomers of average functionality f_avg > 2, Carothers predicts gelation at p_c = 2/f_avg (an overestimate, since the true gel point, from Flory’s statistical theory, comes earlier). Glycerol (f = 3) with phthalic anhydride (f = 2) in the stoichiometric 2:3 ratio has f_avg = (2×3 + 3×2)/(2 + 3) = 2.4 and p_c = 0.833.
5. Step-growth kinetics and the distribution of chain lengths
Esterification is acid-catalysed. For equimolar groups with concentration C = C₀(1 − p) of each, the rate of loss is −dC/dt = k[COOH][OH][catalyst]. Externally catalysed (strong acid added, constant): −dC/dt = k′C², giving 1/C = 1/C₀ + k′t, so Xn = 1 + k′C₀t, linear in time. Self-catalysed (the diacid is the catalyst): −dC/dt = kC³, giving Xn² = 1 + 2kC₀²t, so Xn grows only as t^½ and the uncatalysed reaction is very slow. Plots of 1/(1 − p) against t (catalysed) or 1/(1 − p)² against t (self-catalysed) are straight lines.
Flory most-probable distribution. The probability that a given group has reacted is p, so the number fraction of x-mers is N_x/N = (1 − p)p^(x−1), and the weight fraction is w_x = x(1 − p)²p^(x−1). The averages are Xn = 1/(1 − p), Xw = (1 + p)/(1 − p), so the dispersity Xw/Xn = 1 + p → 2 at high conversion. This value of 2 is a hallmark of linear step-growth polymers formed at equilibrium.
| Feature | Chain-growth (radical) | Step-growth |
|---|---|---|
| Active centres | Only growing chain ends | Any functional group |
| Monomer left at high Mn | Plenty; monomer and polymer coexist | Almost none; monomer gone early |
| Mn versus conversion | High at once, nearly constant | Rises slowly, huge only as p → 1 |
| Long reaction time | Raises yield, not Mn | Raises Mn |
6. Metallocene catalysts and newer (controlled/living) methods
Metallocene catalysts are sandwich complexes of a group-4 metal (Zr, Ti, Hf) between cyclopentadienyl-type rings, activated by a co-catalyst, methylaluminoxane (MAO) (Kaminsky). Because every metal centre is chemically the same (single-site), the polymer has narrow molecular-weight distribution (dispersity near 2, against 4–10 for multi-site Ziegler–Natta), uniform comonomer incorporation (so linear low-density polyethylene with controlled short branching), and tacticity set by the ligand symmetry: bridged (ansa) C₂-symmetric complexes give isotactic and Cs-symmetric ones give syndiotactic polypropylene. Metallocene polyolefins are clearer and tougher at the same density.
Controlled/living radical polymerization keeps chains in a dormant state, with only a tiny fraction active at any moment, so termination is suppressed and all chains grow together. For an ideal living system Xn = [M]₀·p/[I]₀ (p = monomer conversion) and the dispersity approaches 1 + 1/Xn (Poisson). Example: [M]₀ = 2.0 M, [I]₀ = 0.020 M, p = 0.80 gives Xn = 2.0 × 0.80/0.020 = 80 and the chain length grows linearly with conversion, unlike conventional radical polymerization.
| Method | Key idea | Typical use |
|---|---|---|
| ATRP | Alkyl halide initiator, Cu(I)/ligand reversibly abstracts halogen, giving radical and Cu(II)–halide | Block, star, graft architectures of styrenics and (meth)acrylates |
| RAFT | Thiocarbonylthio chain-transfer agent; degenerative transfer exchanges active and dormant chains | Wide monomer range, no metal |
| Ring-opening (ROP) | Strained ring opens: lactide, ε-caprolactone, cyclic ethers, lactams | Polylactide, polycaprolactone, nylon 6 |
| Group-transfer (GTP) | Silyl ketene acetal initiator transfers the silyl group to the incoming (meth)acrylate, nucleophilic catalyst | Living acrylics at room temperature |
Key takeaways
- Xn counts structural units; Mn = Xn·M₀. Functionality above 2 on average leads to branching and gelation at p_c = 2/f_avg (Carothers).
- Free-radical: Rp = k_p[M](f·k_d[I]/k_t)^½, ν = Rp/Ri; Xn = 2ν for combination, ν for disproportionation; Rp ∝ [I]^½, Xn ∝ [I]^−½.
- Mayo: 1/Xn = 1/Xn,0 + C_M + C_S[S]/[M]; transfer shortens chains without changing rate; the gel effect (low k_t) raises rate at high conversion.
- Step-growth: Xn = 1/(1 − p), or (1 + r)/(1 + r − 2rp) with imbalance; high polymer needs p > 0.99; catalysed Xn = 1 + k′C₀t, self-catalysed Xn² = 1 + 2kC₀²t; Xw/Xn = 1 + p.
- Anionic polymerization of clean systems is living (Xn = [M]₀/[I]₀); Ziegler–Natta is multi-site and broad, metallocene single-site and narrow with ligand-controlled tacticity.
- ATRP, RAFT, ROP and GTP give predetermined Xn = [M]₀p/[I]₀, low dispersity and block architectures.
Practice questions (16)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
Which of the following monomers has a functionality of three in condensation polymerization?
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Answer: C — Glycerol
Glycerol has three hydroxyl groups, so f = 3 and it gives branching or networks. Adipic acid, ethylene glycol and hexamethylene diamine each have two reactive groups (f = 2) and give linear chains.In a free-radical polymerization with bimolecular termination, the initiator concentration is doubled at constant temperature and monomer concentration, with no chain transfer. What happens?
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Answer: A — Rate rises by about 1.41 times and Xn falls by the same factor
Rp ∝ [I]^½ and ν ∝ [I]^−½, so both change by √2 in opposite directions. Rate doubling would need first-order dependence on [I], which is what the tempting alternative assumes; it ignores that termination is second order in radicals.In an equimolar step-growth polymerization the extent of reaction is raised from p = 0.95 to p = 0.99. By what factor does the number-average degree of polymerization increase?
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Answer: D — 5 times
Xn = 1/(1 − p): 1/0.05 = 20 and 1/0.01 = 100, a factor of 5. The tiny change in p hides a large change in chain length, which is why the last few percent of conversion matter; the 1.04 ratio of p values is the tempting error.n-Butyllithium initiates styrene in a rigorously dry, impurity-free hydrocarbon solvent. Which description is correct?
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Answer: B — Chains stay active after the monomer is consumed, so a second monomer can add to form a block copolymer
Like-charged carbanion ends repel, and in a pure system no transfer or termination occurs, so the polymerization is living and block copolymers form by adding another monomer. Combination and Xn = 2ν belong to radical polymerization; a Lewis acid co-initiator belongs to cationic polymerization.A mercaptan is added to a free-radical polymerization of styrene as a chain-transfer agent with a large transfer constant. Compared with the run without it, the main effect is
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Answer: C — lower Xn with the rate nearly unchanged
Transfer ends one chain but starts another radical of similar reactivity, so rate stays nearly the same while 1/Xn increases by C_S[S]/[M] in the Mayo equation. Lower rate or a higher Xn would require the new radical to be inactive or the chains not to be cut.In bulk free-radical polymerization of methyl methacrylate, the rate rises sharply at moderate conversion although the initiator is being consumed. The correct explanation is that
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Answer: B — k_t falls as viscous growing chains diffuse slowly, while small monomer still reaches radical ends
This is the gel (Trommsdorff) effect: diffusion-controlled termination slows, radical concentration climbs, and Rp = k_p[M][M•] surges. Monomer concentration actually falls, and viscosity does not make k_d larger; the mechanism stays chain-growth.The principal reason a single-site metallocene catalyst gives polyolefins with narrow molecular-weight distribution is that
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Answer: B — every active metal centre is chemically identical, so all chains grow under the same conditions
Identical sites give a single most-probable distribution with dispersity near 2, while heterogeneous multi-site Ziegler–Natta catalysts superpose several distributions and give a broad one. Methylaluminoxane activates the metal; it does not set chain length.Which pairing of a controlled-polymerization method with its characteristic feature is correct?
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Answer: A — ATRP: a Cu(I)/ligand complex reversibly abstracts a halogen atom from a dormant alkyl halide chain end
ATRP is built on the reversible Cu(I)/Cu(II) halogen exchange. RAFT uses a thiocarbonylthio transfer agent and no metal, nitroxide capping is the nitroxide-mediated method (not group-transfer), and ring-opening uses strained rings such as lactide or caprolactam, not C=C.For a linear polyester made by equimolar step-growth to extent of reaction p = 0.99, the ratio Xw/Xn from the Flory most-probable distribution (Xw = (1 + p)/(1 − p)) is closest to
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Answer: D — 1.99, equal to 1 + p
Xn = 1/(1 − p) = 100 and Xw = 1.99/0.01 = 199, so Xw/Xn = 1 + p = 1.99, the familiar value near 2. A high conversion does not narrow the distribution: both averages grow together.Which of the following statements about free-radical chain polymerization (steady state, bimolecular termination) are correct?
P: High-Mn chains are formed from the very start of the reaction.
Q: The rate of polymerization is proportional to the square root of the initiator concentration.
R: Mn rises steadily and becomes large only as conversion approaches 100%.
S: Chains grow only at the active end, so monomer and high polymer coexist throughout.Show answer
Answer: A — P; B — Q; D — S
P, Q and S describe chain-growth: each chain is completed in about a second, Rp ∝ [I]^½, and unreacted monomer sits beside full-size polymer. R describes step-growth, where Xn = 1/(1 − p) is large only near full conversion.A polyamide is to be made by step-growth from a diacid and a diamine, with Xn of at least 100. Which of the following are required?
P: Extent of reaction of the limiting group above 0.99.
Q: A molar ratio of the two monomers within about 2% of unity.
R: Efficient removal or control of the small-molecule by-product so the equilibrium is not limiting.
S: A small intentional excess of a monofunctional acid to speed the growth.Show answer
Answer: A — P; B — Q; C — R
Xn = 100 needs p ≥ 0.99 at exact stoichiometry (r = 1) and (1 + r)/(1 − r) ≥ 100 means r ≥ 0.98 even at p = 1. By-product removal pushes the equilibrium to high p. A monofunctional acid caps chains and lowers Xn; it never speeds growth.In a free-radical polymerization with no chain transfer, ν is the kinetic chain length (ν = Rp/Ri, with Ri = Rt). Which statements are correct?
P: If termination is entirely by combination, Xn = 2ν.
Q: If termination is entirely by disproportionation, Xn = ν.
R: ν falls when the initiator concentration is raised, at fixed monomer concentration.
S: ν is independent of monomer concentration.Show answer
Answer: A — P; B — Q; C — R
Combination joins two kinetic chains in one molecule (Xn = 2ν), disproportionation leaves one molecule per kinetic chain (Xn = ν), and ν = k_p[M]/(2(f·k_d·k_t[I])^½) falls as [I]^−½. It is directly proportional to [M], so the claim of independence from [M] is wrong.A step-growth polymer is made from A–A and B–B monomers with a stoichiometric ratio r = N_A/N_B = 0.98 (B in 2% excess). The reaction is driven to complete conversion of the A groups (p = 1). Using Xn = (1 + r)/(1 + r − 2rp), calculate the number-average degree of polymerization. Give the answer as an integer.
Numerical answer — type the value.
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Answer: 99
At p = 1: Xn = (1 + 0.98)/(1 + 0.98 − 2 × 0.98) = 1.98/0.02 = 99. A 2% imbalance caps Xn near 100.An equimolar diacid and diol are polymerized to extent of reaction p = 0.99. The mean mass of a structural unit is M₀ = 113 g/mol; neglect end groups. Using Xn = 1/(1 − p), calculate Mn in g/mol. Give the answer as an integer.
Numerical answer — type the value.
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Answer: 11300
Xn = 1/(1 − 0.99) = 100, so Mn = Xn·M₀ = 100 × 113 = 11,300 g/mol.In a free-radical polymerization with no chain transfer, the polymerization rate is Rp = 1.0×10⁻⁴ mol L⁻¹ s⁻¹ and the initiation rate is Ri = 2.0×10⁻⁷ mol L⁻¹ s⁻¹, with steady state (Ri = Rt). Termination is entirely by combination. Calculate Xn. Give the answer as an integer.
Numerical answer — type the value.
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Answer: 1000
ν = Rp/Ri = 1.0×10⁻⁴/2.0×10⁻⁷ = 500. Combination joins two kinetic chains, so Xn = 2ν = 1000.An ideal living polymerization starts with monomer [M]₀ = 2.0 mol/L and initiator [I]₀ = 0.020 mol/L, and every initiator molecule starts one chain with no termination or transfer. Using Xn = [M]₀p/[I]₀, calculate Xn at monomer conversion p = 0.80. Give the answer as an integer.
Numerical answer — type the value.
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Answer: 80
Xn = 2.0 × 0.80/0.020 = 80. In living polymerization Xn grows linearly with conversion.