Polymer Chemistry II: Copolymerization, Reactivity Ratios, Block and Graft Copolymers, Polymerization Techniques

This chapter completes Section XE5.1 “Polymer Chemistry” of XE5 Polymer Science and Engineering. It answers the second half of the heading: “Copolymerization; Monomer reactivity ratios and its significance; Kinetics; Different copolymers; Random, alternating, azeotropic copolymerization; Block and graft copolymers; Techniques for polymerization-bulk, solution, suspension, emulsion.” It teaches the copolymer composition (Mayo–Lewis) equation and how it is derived, the meaning of r₁ and r₂ and the shape of composition curves, random, alternating, ideal and azeotropic behaviour, composition drift and sequence lengths, block and graft copolymers and how they are made, and the four industrial techniques with their advantages, drawbacks and molecular-weight control, including the Smith–Ewart picture of emulsion polymerization. The numericals are instantaneous copolymer composition, azeotropic feed, mean sequence length and the emulsion rate.

1. The copolymer composition equation

When two monomers M₁ and M₂ polymerize together, four propagation steps compete for a chain ending in M₁• or M₂•: M₁• + M₁ (k₁₁), M₁• + M₂ (k₁₂), M₂• + M₂ (k₂₂), M₂• + M₁ (k₂₁). The terminal model assumes that reactivity depends only on the last unit. Monomer reactivity ratios are r₁ = k₁₁/k₁₂ and r₂ = k₂₂/k₂₁: each is the preference of a radical for its own monomer over the other monomer.

Assuming steady state for the two radical types (the cross-propagation rates are equal, k₁₂[M₁•][M₂] = k₂₁[M₂•][M₁]) and long chains, the consumption ratio is the Mayo–Lewis equation:

d[M₁]/d[M₂] = ([M₁]/[M₂]) · (r₁[M₁] + [M₂])/([M₁] + r₂[M₂])

In mole fractions, with f₁ = [M₁]/([M₁] + [M₂]) in the feed and F₁ the mole fraction of M₁ in the polymer formed at that instant:

F₁ = (r₁f₁² + f₁f₂)/(r₁f₁² + 2f₁f₂ + r₂f₂²),
f₂ = 1 − f₁

Worked example. r₁ = 0.4, r₂ = 0.6, f₁ = f₂ = 0.5: numerator = 0.4 × 0.25 + 0.25 = 0.35; denominator = 0.10 + 0.50 + 0.15 = 0.75; F₁ = 0.467. The polymer is slightly poorer in M₁ than the feed because the M₁ radical prefers M₂ (1/r₁ = 2.5) more strongly than the M₂ radical prefers M₁ (1/r₂ = 1.67), so M₂ is built into the chain somewhat faster. Note that this is the instantaneous composition at low conversion, not the average composition of the whole batch.

⚠️ The definition of r
r₁ = k₁₁/k₁₂ puts the homo-propagation constant on top. Inverting it (k₁₂/k₁₁) reverses every conclusion. A monomer pair is also described by two numbers, never one; r₁ > 1 says M₁• prefers M₁, and says nothing about M₂•.

2. Reactivity ratios and what they tell us

Copolymer behaviour from r₁ and r₂
ValuesBehaviourStructure and curve
r₁ = r₂ = 1Ideal random: no preferenceF₁ = f₁ (diagonal)
r₁r₂ = 1 (ideal copolymerization)Both radicals prefer the same monomer equallyRandom, polymer richer in more reactive monomer; curve does not cross the diagonal
r₁ and r₂ both below 1Tendency to alternate; azeotrope existsS-shaped curve crossing the diagonal
r₁ ≈ r₂ ≈ 0Perfectly alternatingF₁ = 0.5 for all feeds (e.g. styrene–maleic anhydride)
r₁ and r₂ both above 1Tendency to block formation (rare)Curve crossing the diagonal, azeotrope unstable
r₁ ≫ 1, r₂ ≪ 1M₁ consumed first, nearly homopolymer thenStrong composition drift

Ideal copolymerization (r₁r₂ = 1): the relative rate at which either radical adds M₁ over M₂ is the same, k₁₁/k₁₂ = k₂₁/k₂₂. The composition equation reduces to F₁ = r₁f₁/(r₁f₁ + f₂); here the polymer is always richer in the more reactive monomer, and the chain is statistically random like the tossing of a biased coin. Alternating copolymers occur when both radicals strongly prefer the other monomer, usually an electron-rich monomer paired with an electron-poor one (a charge-transfer complex or polar effect).

Mean sequence length. The number-average run length of M₁ units in the chain is n₁ = 1 + r₁[M₁]/[M₂]. For r₁ = 3.5 and [M₁]/[M₂] = 2, n₁ = 1 + 3.5 × 2 = 8; for r₁ → 0, n₁ → 1, which is strict alternation.

🎯 Why the polar effect matters
Resonance stabilisation makes the styrene monomer more reactive toward radicals but its own radical far less reactive, and radical stability dominates the cross-propagation rates. In the Q–e scheme of Alfrey and Price, Q measures resonance stabilisation and e the polarity of the double bond; monomers with opposite e values tend to alternate.

3. Azeotropic copolymerization and composition drift

When both r₁ and r₂ are below 1 (or both above 1) the curve of F₁ against f₁ crosses the diagonal. At that feed the polymer has the same composition as the feed, F₁ = f₁, and the feed does not change as the reaction proceeds: an azeotropic copolymerization. Setting F₁ = f₁ in the composition equation gives

f₁(azeotrope) = (1 − r₂)/(2 − r₁ − r₂)

Example: r₁ = 0.3, r₂ = 0.6: f₁ = 0.4/1.1 = 0.364. Feeding the azeotropic mixture makes a polymer of uniform composition from start to finish, which is the standard way to avoid composition drift in a batch.

Composition drift. In any other feed the more reactive monomer is consumed faster, so the feed shifts toward the less reactive monomer and the instantaneous F₁ changes with conversion; a high-conversion product is a blend of chains of different composition. Remedies:
(i) stop at low conversion,
(ii) feed the faster monomer continuously (semi-batch), or
(iii) use the azeotropic feed. The overall copolymer is described by the integrated composition equation, which is not needed numerically here.

🧠 Quick check of the shape
At f₁ = 0 the equation gives F₁ = 0 and at f₁ = 1 gives F₁ = 1. If r₁ > 1 and r₂ < 1, M₁ is favoured everywhere and the curve lies above the diagonal. An azeotrope can exist only when r₁ and r₂ are on the same side of 1.

4. Block and graft copolymers

Statistical (random) and alternating copolymers are decided by the reactivity ratios; block (–AAAA–BBBB–) and graft (a backbone of A carrying branches of B) copolymers are made by deliberate design because their chemically distinct segments are usually immiscible and so separate into nanoscale domains (microphase separation), combining the properties of both parts.

How block and graft copolymers are made
RoutePrincipleExample
Sequential living additionPolymerize A with a living system, then add BAnionic polystyrene–polybutadiene–polystyrene (SBS)
Coupling of end-functional blocksReact preformed telechelic chains, or a step-growthPolyurethane segmented elastomers
Grafting fromRadical sites made on a backbone start side chains, often by chain transfer to polymerStyrene–acrylonitrile on polybutadiene (ABS); polystyrene on rubber in HIPS
Grafting onto / throughReactive end groups attach to a backbone, or a macromonomer is copolymerizedComb or bottle-brush polymers

In SBS the glassy polystyrene end blocks form spheres or cylinders that act as physical crosslinks anchoring rubbery polybutadiene mid-blocks: the material behaves as a rubber at service temperature and flows when heated above the polystyrene Tg, so it is a thermoplastic elastomer that can be moulded and recycled. In ABS and HIPS the rubber phase, grafted so that it bonds to the glassy matrix, toughens an otherwise brittle polymer by promoting crazing and shear yielding.

⚠️ Block is not blend
A block copolymer has covalent links between its segments, so phase separation is limited to nanometre domains; a physical blend of the same two homopolymers separates into micron-scale phases. Both can look cloudy or clear for different reasons, so appearance is a poor test.

5. Polymerization techniques: bulk and solution

Bulk (mass) polymerization has only monomer and a monomer-soluble initiator. Advantages: the purest polymer, simple equipment, high rate, direct casting into shapes (acrylic sheet is cast from partly polymerized methyl methacrylate syrup between plates). Drawbacks: the exotherm is hard to remove because viscosity rises and the gel effect adds more heat, risking runaway, hot spots and a broad molecular-weight distribution; residual monomer is hard to remove. Molecular weight is set by initiator level, temperature and added chain-transfer agent. In step-growth, bulk (melt) polymerization of PET and nylon 66 is routine because the reaction is slower and less exothermic.

Solution polymerization dissolves monomer and initiator in a solvent. Advantages: the solvent absorbs and carries away heat, lowers viscosity (easier stirring, less gel effect), and the solution can be used directly as a lacquer, adhesive or spinning dope. Drawbacks: the solvent adds a chain-transfer route (C_S[S]/[M] in the Mayo equation) and so lowers molecular weight, lowers the monomer concentration and so the rate, and must be recovered, with cost, residue and fire or toxicity concerns. Typical uses: acrylic fibre spinning dope, polyvinyl acetate for adhesives, and coordination polymerization of olefins in hydrocarbon diluent.

6. Suspension and emulsion polymerization

Suspension polymerization disperses water-insoluble monomer as droplets (tens to hundreds of micrometres) in water by stirring, with a polymeric stabiliser such as poly(vinyl alcohol) and a monomer-soluble initiator. Each droplet is a tiny bulk reactor, so kinetics and molecular weight are those of bulk polymerization, while the water carries off the heat. The product is beads that are filtered and dried: PVC, polystyrene (including expandable beads) and PMMA beads. Drawbacks: stabiliser residue, drying, effluent water, and limited use for tacky polymers because droplets stick.

Emulsion polymerization uses water, sparingly soluble monomer, a surfactant above its critical micelle concentration and a water-soluble initiator (persulfate). Monomer sits mostly in large droplets (a reservoir) and in swollen micelles; radicals generated in water enter micelles, which become the loci of polymerization, giving particles of about 0.05 to 0.5 μm. In the Smith–Ewart case 2 limit each particle holds on average n̄ = 0.5 radicals (one radical present half the time, because a second entering radical terminates the first instantly), so

Rp = k_p[M]_p · N · n̄ / N_A,
Xn ∝ k_p[M]_p · N / R_i

with [M]_p the monomer concentration in the particles and N the particle number per litre. Numerical example: k_p = 1000 L mol⁻¹ s⁻¹, [M]_p = 5 mol/L, N = 1×10¹⁷ L⁻¹, n̄ = 0.5, N_A = 6.022×10²³ mol⁻¹ give Rp = 1000 × 5 × 10¹⁷ × 0.5/6.022×10²³ = 4.15×10⁻⁴ mol L⁻¹ s⁻¹. Because radicals are isolated in separate particles, termination is slowed while the rate scales with N, so emulsion polymerization gives high rate and high molecular weight together; the rate and Xn rise with surfactant concentration (more particles).

The four techniques compared
FeatureBulkSolutionSuspensionEmulsion
MediumMonomer onlyOrganic solventWater, dropletsWater, micelles
InitiatorMonomer-solubleSoluble in solutionMonomer-solubleWater-soluble
Heat removalPoorGoodVery goodVery good
PurityHighestSolvent residueStabiliser residueSurfactant residue
Product formCast block or meltSolutionBeadsLatex, 0.05–0.5 μm
⚠️ Initiator solubility is the clue
Suspension: initiator inside the droplet, so it behaves like many small bulk reactors. Emulsion: initiator in the water, radicals captured by micelles. Questions on which technique gives high rate and high Mn together hinge on this.

Key takeaways

  • Mayo–Lewis: F₁ = (r₁f₁² + f₁f₂)/(r₁f₁² + 2f₁f₂ + r₂f₂²), with r₁ = k₁₁/k₁₂ and r₂ = k₂₂/k₂₁; it gives the instantaneous composition at low conversion.
  • r₁r₂ = 1 is ideal random; r₁ ≈ r₂ ≈ 0 alternating; both below 1 an azeotrope at f₁ = (1 − r₂)/(2 − r₁ − r₂); both above 1 a block tendency.
  • Away from the azeotrope the feed drifts as the more reactive monomer is used up; control drift by low conversion, semi-batch feed or azeotropic feed. Mean M₁ run length is 1 + r₁[M₁]/[M₂].
  • Block copolymers come from sequential living polymerization or coupling, graft copolymers from chain transfer or macromonomers; SBS is a thermoplastic elastomer, ABS and HIPS are rubber-toughened.
  • Bulk is purest but hard to cool; solution cools easily but lowers Mn by transfer and needs solvent recovery; suspension gives beads, each droplet a bulk reactor; emulsion gives latex with high rate and high Mn together.
  • Smith–Ewart case 2: Rp = k_p[M]_p N n̄/N_A with n̄ = 0.5; raising surfactant raises N, so rate and Xn rise together.

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.

  1. In the terminal model of copolymerization, the monomer reactivity ratio r₁ is defined as

    1. k₁₂/k₁₁, the ratio of cross- to homo-propagation for the M₁ radical
    2. k₁₁/k₁₂, the ratio of homo- to cross-propagation for the M₁ radical
    3. k₂₂/k₂₁, the ratio for the M₂ radical
    4. k₁₁/k₂₂, the ratio of the two homo-propagation constants
    Show answer

    Answer: B — k₁₁/k₁₂, the ratio of homo- to cross-propagation for the M₁ radical

    r₁ = k₁₁/k₁₂ measures how strongly an M₁-ended chain prefers adding M₁ rather than M₂. The inverted ratio reverses the meaning, k₂₂/k₂₁ is r₂, and k₁₁/k₂₂ compares different radicals and is not a reactivity ratio.

  2. Styrene and maleic anhydride copolymerize with both reactivity ratios close to zero. The copolymer formed is

    1. alternating, with F₁ near 0.5 over a wide range of feed compositions
    2. a block copolymer with long runs of each unit, since like units are preferred
    3. a random copolymer whose composition equals the feed
    4. a mixture of the two homopolymers, since neither radical adds the other monomer
    Show answer

    Answer: A — alternating, with F₁ near 0.5 over a wide range of feed compositions

    With r₁ ≈ r₂ ≈ 0, each radical adds only the other monomer, so units alternate and the polymer is about 50:50 whatever the feed. A random copolymer equal to the feed needs r₁ = r₂ = 1; homopolymers or blocks need r values above 1.

  3. For an ideal copolymerization with r₁ = 2 and r₂ = 0.5, an equimolar feed gives a copolymer that is

    1. richer in M₂, since M₂ has the smaller reactivity ratio
    2. equal to the feed, as in an azeotrope
    3. strictly alternating because r₁r₂ is near unity
    4. richer in M₁, with F₁ ≈ 0.67
    Show answer

    Answer: D — richer in M₁, with F₁ ≈ 0.67

    r₁r₂ = 1 is ideal: F₁ = r₁f₁/(r₁f₁ + f₂) = 1/(1 + 0.5) = 0.67, so the polymer is richer in the more reactive M₁. An ideal system has no azeotrope (the curve does not cross the diagonal) and is random, not alternating.

  4. A batch copolymerization has r₁ = r₂ = 0.5, so the azeotrope is at f₁ = 0.5. It is started with a feed of f₁ = 0.8 and run to moderate conversion. What happens to the monomer feed composition as it proceeds?

    1. f₁ falls toward 0.5, because the polymer formed at the start has F₁ ≈ 0.73, lower than f₁
    2. f₁ rises toward 1, because M₁ is the majority monomer and is used up more slowly
    3. f₁ stays at 0.8, because the polymer composition equals the feed at every conversion
    4. f₁ falls to zero, because M₂ is the more reactive monomer and is used last
    Show answer

    Answer: A — f₁ falls toward 0.5, because the polymer formed at the start has F₁ ≈ 0.73, lower than f₁

    At f₁ = 0.8: numerator 0.5 × 0.64 + 0.16 = 0.48, denominator 0.32 + 0.32 + 0.02 = 0.66, F₁ = 0.727. The polymer takes less M₁ than its share, so the feed loses M₁ and f₁ drifts toward the stable azeotrope at 0.5 and not beyond it.

  5. Polystyrene–polybutadiene–polystyrene (SBS) behaves as a rubber at room temperature and can be moulded when heated. The reason is that

    1. covalent sulfur crosslinks form during moulding
    2. the polybutadiene blocks crystallize and melt on heating
    3. the glassy polystyrene domains act as physical crosslinks that soften above their Tg
    4. it is a blend of polystyrene and rubber that is mixed at the molecular level and so melts at one temperature
    Show answer

    Answer: C — the glassy polystyrene domains act as physical crosslinks that soften above their Tg

    Immiscible blocks microphase-separate; polystyrene end blocks form glassy domains that tie rubbery mid-blocks together, and when heated above the polystyrene Tg the network flows, so it is a thermoplastic elastomer. Vulcanization (sulfur) is irreversible and would prevent remoulding.

  6. Which route is used to make the toughened grade high-impact polystyrene (HIPS)?

    1. Polymerizing styrene in the presence of dissolved polybutadiene so that some styrene grafts onto the rubber
    2. Step-growth reaction of styrene with a diol in the melt, giving a polyester
    3. Anionic polymerization of styrene to a living chain, then termination of the chain ends with water
    4. Cationic polymerization of styrene with BF₃ in water at elevated temperature
    Show answer

    Answer: A — Polymerizing styrene in the presence of dissolved polybutadiene so that some styrene grafts onto the rubber

    Radical transfer to polybutadiene creates graft sites; the grafted rubber particles bond to the glassy polystyrene matrix and toughen it. The other routes give plain polystyrene or have no rubber to graft onto.

  7. A polymerization uses water, a surfactant above its critical micelle concentration, a water-soluble persulfate initiator and a sparingly soluble monomer, and gives a latex of 0.1 μm particles. This is

    1. bulk polymerization
    2. solution polymerization
    3. suspension polymerization
    4. emulsion polymerization
    Show answer

    Answer: D — emulsion polymerization

    Surfactant micelles, a water-soluble initiator and sub-micrometre latex particles define emulsion polymerization. Suspension uses a monomer-soluble initiator and stabiliser and gives beads of tens to hundreds of micrometres.

  8. Compared with the same reaction in bulk, solution polymerization of a vinyl monomer typically gives

    1. a higher molecular weight because the solvent protects the radicals from termination
    2. a more severe gel effect because the solvent raises the viscosity of the reacting medium
    3. a lower rate and a lower molecular weight, through dilution and chain transfer to the solvent
    4. a polymer free of any residue because the solvent evaporates completely on drying
    Show answer

    Answer: C — a lower rate and a lower molecular weight, through dilution and chain transfer to the solvent

    Dilution lowers [M] and so Rp, and transfer to solvent adds C_S[S]/[M] to 1/Xn. The solvent lowers viscosity and thus eases heat removal and weakens the gel effect, and complete solvent removal is difficult.

  9. In an emulsion polymerization following Smith–Ewart case 2 (n̄ = 0.5), the surfactant concentration is increased so that the number of latex particles per litre doubles, with the same initiation rate and monomer concentration in the particles. The polymerization rate and the number-average degree of polymerization change by factors of

    1. 2 and 2
    2. 2 and 1 (unchanged)
    3. √2 and 1/√2
    4. 1 (unchanged) and 2
    Show answer

    Answer: A — 2 and 2

    Rp = k_p[M]_p N n̄/N_A is proportional to N, and Xn ∝ k_p[M]_p N/R_i also doubles at fixed R_i, since each radical now gets more monomer before the next radical enters. Emulsion polymerization therefore raises rate and molecular weight together, unlike bulk, where they trade off (rate ∝ [I]^½, Xn ∝ [I]^−½).

  10. Which of the following statements on binary copolymerization are correct?
    P: An azeotropic composition exists only when r₁ and r₂ are both below 1 or both above 1.
    Q: With r₁ = 0.2 and r₂ = 5, the polymer is strictly alternating.
    R: With r₁ = r₂ = 0, the polymer composition is F₁ = 0.5 for every feed composition.
    S: A value r₁ > 1 shows that the M₂-ended radical also prefers to add M₁.

    1. P
    2. Q
    3. R
    4. S
    Show answer

    Answer: A — P; C — R

    The azeotrope formula f₁ = (1 − r₂)/(2 − r₁ − r₂) lies between 0 and 1 only when r₁ and r₂ are on the same side of 1, and r₁ = r₂ = 0 forces alternation. With r₁ = 0.2 and r₂ = 5, M₂ radicals strongly prefer M₂, giving long M₂ runs, not alternation. r₁ describes only the M₁-ended radical; r₂ describes the other.

  11. Which of the following statements about bulk free-radical polymerization are correct?
    P: It gives the purest polymer, with no solvent or dispersant to remove.
    Q: Heat removal is easy because the viscosity stays low throughout.
    R: The gel effect can increase the rate and the heat release at high conversion.
    S: Monomer can be polymerized directly in a mould or between plates to give a cast shape.

    1. P
    2. Q
    3. R
    4. S
    Show answer

    Answer: A — P; C — R; D — S

    Bulk polymerization has no extra components, casts directly (acrylic sheet), but viscosity climbs and the gel effect worsens heat removal, so the statement on easy heat removal is the reverse of the truth.

  12. Which of the following statements about emulsion polymerization are correct?
    P: The initiator is water-soluble, so radicals are generated in the aqueous phase.
    Q: The polymerization rate rises with the number of latex particles.
    R: Several radicals share each growing particle, so termination is fast and Xn is low.
    S: The large monomer droplets mainly act as reservoirs that feed monomer to the polymer particles.

    1. P
    2. Q
    3. R
    4. S
    Show answer

    Answer: A — P; B — Q; D — S

    Persulfate radicals form in water, enter micelles, and each particle then contains on average about half a radical, so termination is slow and Xn is high; rate ∝ N. The droplets are monomer reservoirs. The statement that termination is fast and Xn low is the reverse of the true compartmentalisation effect.

  13. A copolymerization has r₁ = 0.4 and r₂ = 0.6 (terminal model, r₁ = k₁₁/k₁₂, r₂ = k₂₂/k₂₁). The feed is equimolar, f₁ = f₂ = 0.5. Using F₁ = (r₁f₁² + f₁f₂)/(r₁f₁² + 2f₁f₂ + r₂f₂²), calculate the instantaneous mole fraction of M₁ in the copolymer, to three decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 0.467

    Numerator = 0.4 × 0.25 + 0.25 = 0.35; denominator = 0.10 + 0.50 + 0.15 = 0.75; F₁ = 0.35/0.75 = 0.4667, i.e. 0.467.

  14. For a copolymerization with r₁ = 0.3 and r₂ = 0.6, an azeotrope exists. Using the condition F₁ = f₁, calculate the mole fraction f₁ of M₁ in the feed at the azeotrope, to three decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 0.364

    f₁ = (1 − r₂)/(2 − r₁ − r₂) = 0.4/(2 − 0.9) = 0.4/1.1 = 0.3636, i.e. 0.364.

  15. In a copolymerization with r₁ = 3.5, the feed has [M₁]/[M₂] = 2. Using the number-average sequence length n₁ = 1 + r₁[M₁]/[M₂] of M₁ units, calculate n₁. Give the answer as an integer.

    Numerical answer — type the value.

    Show answer

    Answer: 8

    The mean run of M₁ units is the reciprocal of the chance that an M₁-ended chain adds M₂: n₁ = 1 + r₁[M₁]/[M₂] = 1 + 3.5 × 2 = 1 + 7 = 8. For r₁ → 0 the run length falls to 1 (alternation).

  16. In an emulsion polymerization (Smith–Ewart case 2, n̄ = 0.5) the propagation constant is k_p = 1000 L mol⁻¹ s⁻¹, the monomer concentration in the particles is [M]_p = 5 mol/L and the particle number is N = 1×10¹⁷ per litre. Using Rp = k_p[M]_p N n̄/N_A with N_A = 6.022×10²³ mol⁻¹, calculate Rp in units of 10⁻⁴ mol L⁻¹ s⁻¹, to two decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 4.15

    Rp = 1000 × 5 × 1×10¹⁷ × 0.5/6.022×10²³ = 2.5×10²⁰/6.022×10²³ = 4.151×10⁻⁴ mol L⁻¹ s⁻¹, i.e. 4.15 in the stated units.