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Chapter 8 • Theory & Derivations

Ionic & Living Polymerization: Mechanisms, Stereospecificity & Catalysis

Ionic chain polymerization fundamentals, electronic selectivity of monomers, anionic initiation via organolithium and sodium naphthalenide electron transfer, Michael Szwarc's discovery of living polymers, narrow Poisson distributions, Winstein ion-pair spectrum (free ions, solvent-separated pairs, contact pairs), cationic polymerization mechanisms with Lewis acid co-initiators, stereospecific coordination catalysis via heterogeneous Ziegler-Natta and homogeneous metallocenes, and modern controlled radical architectures (ATRP, RAFT).

§8.1 Fundamentals of Ionic Polymerization: Electronic Selectivity & Counterion Cages

Unlike free-radical polymerizations where unpaired electrons are electrically neutral and largely unperturbed by solvent polarity, ionic polymerizations propagate via charged ionic centers—either carbanions (anionic) or carbocations (cationic).

Monomer Selectivity and Electronic Substituent Effects

The ability of an alkene monomer ($CH_2=CHX$) to polymerize via ionic mechanisms is strictly dictated by the electronic resonance and inductive properties of the pendant substituent $X$:

1. Monomers for Anionic Polymerization:

  • Require electron-withdrawing substituents ($-CN, -NO_2, -COOR, -COCH_3$, or aromatic rings) that stabilize the developing negative charge on the propagating carbanion:
  • Examples: Acrylonitrile, methyl methacrylate, nitroethylene, vinylidene cyanide, styrene, and 1,3-dienes.

2. Monomers for Cationic Polymerization:

  • Require electron-donating substituents ($-OR, -NR_2, -OH$, alkyl groups, or phenyl) that stabilize the positive charge on the propagating carbocation via induction or lone-pair resonance:
  • Examples: Vinyl ethers ($CH_2=CHOR$), isobutylene ($CH_2=C(CH_3)_2$), $lpha$-methylstyrene, $N$-vinylcarbazole, and cyclic ethers (tetrahydrofuran, oxirane).

Counterions and the Ionic Solvation Cage

Because macroscopic electroneutrality must be strictly preserved, every active propagating ionic chain end is accompanied by an oppositely charged counterion (gegenion):

  • In anionic polymerization: A propagating carbanion ($-C^-$) is paired with a metal cation ($Li^+, Na^+, K^+, Cs^+$).
  • In cationic polymerization: A propagating carbocation ($-C^+$) is paired with a non-nucleophilic complex counter-anion ($BF_3OH^-, AlCl_4^-, SbCl_6^-, PF_6^-$).

The counterion remains in close electrostatic proximity to the active center, strongly modulating the propagation rate, activation energy, and stereochemical insertion geometry.

§8.2 Anionic Polymerization: Organolithium Reagents & Electron Transfer Mechanisms

1. Direct Nucleophilic Addition (Alkyllithium Initiators)

Alkyllithium reagents (such as $n$-butyllithium, $s$-butyllithium, or $t$-butyllithium) are the most widely employed anionic initiators in non-polar hydrocarbon solvents:

\[R^-\text{Li}^+ + \text{CH}_2=\text{CHX} \xrightarrow{k_i} R-\text{CH}_2-\text{C}^-\text{HX}\ \text{Li}^+\]

Initiation efficiency is strongly influenced by the degree of alkyllithium aggregation in solution:

  • In hydrocarbon solvents (cyclohexane, benzene, hexane), $n$-butyllithium forms hexamers $(n-\text{BuLi})_6$, while $s$-butyllithium forms tetramers $(s-\text{BuLi})_4$. Only the tiny dissociated unimer fraction undergoes initiation, resulting in slow initiation relative to propagation.
  • Adding trace amounts of coordinating polar Lewis bases (such as tetrahydrofuran, THF, or diethyl ether) de-aggregates the lithium clusters into solvated monomers, accelerating initiation by several orders of magnitude.

2. Electron Transfer Initiation (Sodium Naphthalenide)

In 1956, Michael Szwarc revolutionized polymer science by demonstrating initiation via one-electron transfer using sodium naphthalenide in THF:

  1. Sodium metal dissolves in a THF solution of naphthalene to form the dark green sodium naphthalenide radical-anion:
\[\text{Na} + \text{Naphthalene} \xrightleftharpoons{\text{THF}} \text{Na}^+ + [\text{Naphthalene}]^{\bullet-}\]
  1. When styrene is added, the naphthalene radical-anion transfers an electron to the styrene double bond, regenerating neutral naphthalene and creating a styrene radical-anion:
\[[\text{Naphthalene}]^{\bullet-} + \text{CH}_2=\text{CHPh} \xrightarrow{} \text{Naphthalene} + [^\bullet\text{CH}_2-\text{C}^-\text{HPh}]\ \text{Na}^+\]
  1. Within microseconds, two styrene radical-anions undergo instantaneous head-to-head coupling of their radical ends, while their anionic carbanionic ends remain intact:
\[2 [^\bullet\text{CH}_2-\text{C}^-\text{HPh}]\ \text{Na}^+ \xrightarrow{\text{coupling}} \text{Na}^+\ [^-\text{CHPh}-\text{CH}_2-\text{CH}_2-\text{CHPh}^-]\ \text{Na}^+\]

This produces a dianionic living telechelic chain capable of propagating symmetrically in both directions simultaneously!

§8.3 Living Polymerization Principles: Michael Szwarc Discovery & Absence of Termination

Szwarc's Breakthrough Discovery (1956)

Michael Szwarc demonstrated that in an ultra-pure, aprotic, deoxygenated environment (rigorously free of moisture, air, and electrophilic impurities), an anionic polymer solution of polystyrene in THF exhibits a persistent red-orange color attributable to active benzylic carbanions ($-\text{CH}_2-\text{C}^-\text{HPh}$). When all monomer is consumed, the color does not fade, and the viscosity of the solution remains constant for weeks or months. Upon introducing a fresh batch of styrene or another compatible monomer (such as isoprene or methyl methacrylate):

  • Polymerization resumes immediately.
  • The molecular weight of the chains increases in direct proportion to the added monomer.
  • No new chains are formed; existing chains simply resume growth.

Szwarc termed these macromolecular systems living polymers.

Criteria for a Truly Living Polymerization

A chain polymerization is classified as strictly 'living' if it satisfies the following diagnostic criteria:

1. Absence of Spontaneous Termination and Chain Transfer:

The concentration of active propagating chain ends $[P^*]$ remains constant throughout the entire course of reaction:

\[[P^*] = [P^*]_0 = \text{constant}\]

2. First-Order Kinetic Linear Rate:

A plot of $\ln([M]_0 / [M])$ versus reaction time $t$ is strictly linear:

\[-\frac{d[M]}{dt} = k_p [P^*] [M] \implies \ln\left( \frac{[M]_0}{[M]} \right) = k_p [P^*] t\]

3. Linear Growth of Molecular Weight with Conversion:

The number-average degree of polymerization $X_n$ increases in direct linear proportion to monomer conversion $p$:

\[X_n = \frac{[M]_0 - [M]}{[I]_0} = p \frac{[M]_0}{[I]_0}\]

4. Near-Monodisperse Molecular Weight Distribution:

The resulting polymer possesses an exceptionally narrow Poisson distribution where the polydispersity index approaches unity:

\[\text{Đ} = \frac{M_w}{M_n} \approx 1.0 + \frac{1}{X_n} \to 1.00\]

5. Chain-End Telechelic Functionalization & Block Copolymerization:

Sequential addition of a second monomer yields cleanly defined $A-B$ or $A-B-A$ block copolymers with near-$100\%$ block efficiency.

§8.4 Molecular Weight Control & Poisson Distribution in Living Systems

Derivation of the Poisson Distribution

Consider a living polymerization system where:

  1. All initiator molecules initiate chains simultaneously at $t = 0$ ($k_i \gg k_p$).
  2. All active chains have an identical, invariant probability of adding monomer.
  3. Spontaneous termination and chain transfer are completely absent.

Let $N$ be the number of monomer molecules consumed per active chain center, so the average degree of polymerization is:

\[\bar{\nu} = X_n - 1 = \frac{[M]_0 - [M]}{[I]_0}\]

According to Flory's statistical derivation, the probability $P(x)$ that a chain has added exactly $x$ monomer units follows the Poisson distribution:

\[P(x) = \frac{\bar{\nu}^x \exp(-\bar{\nu})}{x!}\]

Molecular Weight Moments

The number-average degree of polymerization is:

\[X_n = 1 + \bar{\nu} = 1 + \frac{[M]_0 - [M]}{[I]_0}\]

The weight-average degree of polymerization is obtained from the second moment of the Poisson distribution:

\[X_w = 1 + \bar{\nu} + \frac{\bar{\nu}}{1 + \bar{\nu}}\]

The polydispersity index (dispersity $\text{Đ}$) is:

\[\text{Đ} = \frac{X_w}{X_n} = \frac{1 + \bar{\nu} + \frac{\bar{\nu}}{1 + \bar{\nu}}}{1 + \bar{\nu}} = 1 + \frac{\bar{\nu}}{(1 + \bar{\nu})^2} = 1 + \frac{X_n - 1}{X_n^2} \approx 1 + \frac{1}{X_n}\]

Physical Implications of Poisson Dispersity

  • For $X_n = 50$: $\text{Đ} = 1 + 1/50 = 1.020$.
  • For $X_n = 100$: $\text{Đ} = 1 + 1/100 = 1.010$.
  • For $X_n = 1,000$: $\text{Đ} = 1 + 1/1000 = 1.001$.

In contrast to step-growth ($ ext{Đ} = 2.0$) or radical polymerization ($ ext{Đ} = 1.5 - 2.0$), living polymerization produces polymers that are essentially monodisperse.

§8.5 Ion-Pair Equilibria: Free Ions, Loose/Solvent-Separated Pairs & Contact Pairs

In ionic polymerizations, the propagating active center does not exist as a single unique chemical species. Instead, it participates in a dynamic thermodynamic equilibrium between multiple states of ionic association, known as the Winstein ion-pair spectrum:

\[(P^-\ M^+)_n \xrightleftharpoons{K_{\text{assoc}}} P^-\ M^+ \xrightleftharpoons{K_s} P^- \parallel M^+ \xrightleftharpoons{K_d} P^- + M^+\]

1. Aggregated Multimers $((P^- M^+)_n)$:

Predominant in non-polar hydrocarbon solvents (e.g., cyclohexane, heptane). Organolithium chain ends associate into dimers $(P^- Li^+)_2$. Aggregates are dormant and do not propagate ($k_{p, \text{agg}} = 0$).

2. Contact (Tight / Intimate) Ion Pairs $(P^- M^+)$:

The carbanion and counterion are in direct van der Waals contact with no intervening solvent molecules. The electrostatic coulombic attraction is strong ($E_{\text{coul}} \sim 150\text{ kJ/mol}$). Propagation rate constant is modest ($k_{p, \pm} \approx 10^1 - 10^2\text{ L/(mol s)}$).

3. Solvent-Separated (Loose) Ion Pairs $(P^- \parallel M^+)$:

One or more solvent molecules intercalate between the carbanion and counterion, screening electrostatic attraction. The reactivity of loose ion pairs is much higher ($k_{p, s} \approx 10^4 - 10^5\text{ L/(mol s)}$).

4. Free Carbanions $(P^-)$:

The ion pair is completely dissociated into solvated independent ions. Unshielded by counterion electrostatic hindrance, free ions propagate with colossal velocity:

\[k_{p, -} \approx 10^5 - 10^6\text{ L/(mol s)}\]

The Apparent Propagation Rate Constant $k_p^{\text{app}}$

Under typical conditions in polar solvents like THF, an equilibrium exists between contact/solvent-separated ion pairs $(P^-\ M^+)$ and free ions $(P^-)$ governed by dissociation constant $K_d$:

\[P^-\ M^+ \xrightleftharpoons{K_d} P^- + M^+ \quad \implies K_d = \frac{[P^-] [M^+]}{[P^-\ M^+]} = \frac{\alpha^2 c}{1 - \alpha} \approx \alpha^2 c\]

where $c$ is the total concentration of living chain ends and $\alpha$ is the degree of dissociation ($lpha = \sqrt{K_d / c}$). The experimentally observed overall propagation rate constant $k_p^{\text{app}}$ is:

\[k_p^{\text{app}} = (1 - \alpha) k_{p, \pm} + \alpha k_{p, -} \approx k_{p, \pm} + k_{p, -} \frac{K_d^{1/2}}{c^{1/2}}\]

A plot of $k_p^{\text{app}}$ versus $1/\sqrt{c}$ yields a straight line where:

  • Intercept $= k_{p, \pm}$ (ion-pair propagation rate constant).
  • Slope $= k_{p, -} K_d^{1/2}$.

Even though free ions constitute less than $1\%$ of total active centers in THF ($K_d \sim 10^{-7}\text{ M}$), because $k_{p, -} \approx 1,000 \times k_{p, \pm}$, free ions account for over $90\%$ of all monomer consumption!

§8.6 Cationic Polymerization: Carbocation Intermediates, Lewis Acids & Chain Transfer

Active Centers and Reaction Conditions

Cationic polymerization propagates via electron-deficient carbocations (carbonium / carbenium ions, $-CH_2-C^+HR$). Because carbocations undergo extremely rapid unimolecular side reactions—$eta$-hydride elimination, hydride shift rearrangements, and chain transfer to monomer—conventional cationic polymerization must typically be performed at ultra-low temperatures ($-80^\circ\text{C}$ to $-100^\circ\text{C}$) to suppress transfer and achieve high molecular weight.

Initiation: The Lewis Acid / Co-initiator Synergy

True initiator systems require a binary combination of a strong Lewis acid (Friedel-Crafts catalyst) and a proton donor (the 'co-initiator'):

  • Lewis acids: $BF_3, AlCl_3, SnCl_4, TiCl_4$.
  • Co-initiators (Brønsted acids / protogens): Trace $\text{H}_2\text{O}, ROH, HCl$.

For example, in the industrial synthesis of polyisobutylene (butyl rubber):

\[BF_3 + \text{H}_2\text{O} \xrightleftharpoons{} H^+ [BF_3OH]^-\]

The generated superacid proton transfers to the isobutylene double bond to form a stable tertiary carbocation:

\[H^+ [BF_3OH]^- + \text{CH}_2=\text{C(CH}_3)_2 \xrightarrow{} (\text{CH}_3)_3\text{C}^+\ [BF_3OH]^-\]

Propagation and Chain Transfer to Monomer

Propagation occurs by electrophilic addition:

\[\sim\text{CH}_2-\text{C}^+(\text{CH}_3)_2 + \text{CH}_2=\text{C(CH}_3)_2 \xrightarrow{k_p} \sim\text{CH}_2-\text{C(CH}_3)_2-\text{CH}_2-\text{C}^+(\text{CH}_3)_2\]

The dominant termination pathway is chain transfer to monomer via $\beta$-proton transfer:

\[\sim\text{CH}_2-\text{C}^+(\text{CH}_3)_2 + \text{CH}_2=\text{C(CH}_3)_2 \xrightarrow{k_{tr,M}} \sim\text{CH}=\text{C(CH}_3)_2 + (\text{CH}_3)_3\text{C}^+\]

Because this transfer produces an identical new initiating carbocation, the kinetic chain continues while the individual macromolecule is terminated. The ratio of transfer to propagation is governed by the difference in activation energies:

\[\ln\left( \frac{k_{tr,M}}{k_p} \right) = \ln\left( \frac{A_{tr}}{A_p} \right) - \frac{E_{a, tr} - E_{a, p}}{R T}\]

Because $E_{a, tr} > E_{a, p}$, lowering temperature decreases $k_{tr,M} / k_p$, which is why industrial butyl rubber polymerization is operated cryogenically at $-100^\circ\text{C}$ in liquid methyl chloride.

§8.7 Coordination Polymerization: Ziegler-Natta Catalysts & Stereospecificity

In 1953–1954, Karl Ziegler and Giulio Natta discovered coordination polymerization catalysts, winning the 1963 Nobel Prize in Chemistry for synthesizing linear unbranched polyethylene and stereoregular crystalline polyolefins.

Heterogeneous Ziegler-Natta Catalysts

A classic Ziegler-Natta catalyst is formed by reacting a transition-metal halide from Groups 4–8 with an organometallic alkylating compound from Groups 1–3:

  • Typical system: Titanium tetrachloride or titanium trichloride ($\text{TiCl}_4$ or $\gamma-\text{TiCl}_3$) combined with triethylaluminum ($\text{Al(C}_2\text{H}_5)_3$, $\text{AlEt}_3$).
  • Modern supported catalysts: $\text{TiCl}_4$ supported on activated magnesium chloride ($\text{MgCl}_2$) with internal/external electron donors (phthalates, silanes).

The Cossee-Arlman Mechanism of Monomer Insertion

Chain growth in coordination polymerization proceeds via the Cossee-Arlman mechanism at an octahedral titanium active center on the crystal surface:

  1. The titanium atom has an active metal-carbon $\sigma$-bond ($Ti-P$) connected to the growing polymer chain, and an adjacent vacant coordination site ($\Box$).
  2. A monomer molecule (ethylene or propylene) coordinates to the vacant site via its $\pi$-electrons, forming a titanium-olefin $\pi$-complex.
  3. Four-Center Migratory Insertion: The coordinated monomer inserts into the polarized $Ti-P$ bond via a four-membered cyclic transition state:
\[Ti - \text{CH}_2 - \text{CH}_2 - P\]
  1. The growing polymer chain migrates to the position previously occupied by the coordinated monomer, regenerating a vacant coordination site at the original chain position.

Stereochemical Control of Polypropylene

When propylene ($CH_2=CH(CH_3)$) polymerizes:

  • Heterogeneous Ziegler-Natta catalysts: The asymmetric steric environment created by surrounding bridging chlorine atoms on the rigid crystal surface forces every incoming propylene monomer to coordinate in the identical enantiomorphic orientation, producing isotactic polypropylene ($i\text{-PP}$, crystalline, $T_m \approx 165^\circ\text{C}$).
  • In contrast, uncoordinated free-radical polymerization yields amorphous, gummy atactic polypropylene ($a\text{-PP}$) with no engineering utility.

§8.8 Metallocene Catalysts & Modern Controlled Architectures (ATRP, RAFT)

Homogeneous Metallocene Catalysts (Single-Site Catalysis)

In the 1980s, Walter Kaminsky and Hansjörg Sinn discovered that combining metallocene complexes—bis(cyclopentadienyl) zirconium or titanium dichlorides ($Cp_2ZrCl_2$)—with methylaluminoxane (MAO, $[-Al(CH_3)-O-]_n$) creates soluble, ultra-active homogeneous catalysts.

  • Single-Site Nature: Unlike multi-site heterogeneous catalysts which produce broad molecular weight distributions ($\text{Đ} \approx 4 - 8$), metallocenes have identical, well-defined molecular coordination environments, yielding polyolefins with uniform narrow distributions ($\text{Đ} \approx 2.0$) and uniform comonomer incorporation (LLDPE).
  • Chiral Metallocenes (Brintzinger Complexes): Ansa-metallocenes with bridged indenyl ligands ($C_2$-symmetric bridged catalysts) synthesize ultra-pure isotactic polypropylene, while $C_s$-symmetric catalysts synthesize syndiotactic polypropylene ($s\text{-PP}$).

Modern Controlled Radical Polymerization (CRP / RDRP)

While living ionic polymerizations offer perfect molecular weight control, they require rigorous air/moisture-free conditions and are incompatible with polar monomers (acrylic acid, hydroxyethyl methacrylate). Reversible-Deactivation Radical Polymerizations (RDRP) achieve living character in radical systems through dynamic equilibria between active radicals and dormant species:

1. Atom Transfer Radical Polymerization (ATRP) (Krzysztof Matyjaszewski, 1995):

A transition-metal complex ($Cu^I X / L$) abstracts a halogen atom ($X$) from an alkyl halide dormant chain ($P_n-X$) via reversible one-electron redox transfer:

\[P_n-X + Cu^I/L \xrightleftharpoons[k_{\text{deact}}]{k_{\text{act}}} P_n^\bullet + Cu^{II}X/L\]

The equilibrium heavily favors the dormant state ($K_{\text{ATRP}} = k_{\text{act}} / k_{\text{deact}} \approx 10^{-7} - 10^{-4}$). The radical concentration is kept ultralow ($[P_n^\bullet] \sim 10^{-8}\text{ M}$), virtually eliminating bimolecular termination.

2. Reversible Addition-Fragmentation Chain Transfer (RAFT) (Ezio Rizzardo, Graeme Moad, San Thang, 1998):

Employs a thiocarbonylthio chain transfer agent ($S=C(Z)-S-R$, such as dithiobenzoates or trithiocarbonates) to degenerate chain growth via reversible addition-fragmentation equilibria, allowing synthesis of complex block, star, and brush architectures under conventional radical conditions.

Worked Practice Problems (9 Challenge Exercises)

Multi-step solved problems covering end-to-end vector statistics, radius of gyration, persistence length, characteristic ratio, and tacticity stereochemistry with line-by-line mathematical proofs.

foundation Example 8.1: Living Anionic Polymerization $M_n$ Prediction and Re-Initiation

A living anionic polymerization of styrene is performed in anhydrous cyclohexane at $40.0^\circ\text{C}$ initiated with $s$-butyllithium ($s-\text{BuLi}$, formula weight $64.06\text{ g/mol}$). The reactor is charged with $V = 2.00\text{ L}$ of cyclohexane, $m_{\text{styrene}} = 208.3\text{ g}$ of styrene monomer ($M_0 = 104.15\text{ g/mol}$), and $n_{\text{init}} = 4.00 \times 10^{-3}\text{ moles}$ of $s-\text{BuLi}$. (a) Assuming initiation is complete and instantaneous ($f = 1.0$) with zero chain transfer or termination, calculate the number-average molecular weight $M_n$ at $100\%$ conversion. (b) After all styrene is consumed, an aliquot of the living polymer is analyzed and gives $M_n = 52,100\text{ g/mol}$. A second charge of isoprene ($m_{\text{isoprene}} = 136.2\text{ g}$, $M_{\text{iso}} = 68.12\text{ g/mol}$) is injected into the living polystyrene solution. Calculate the number-average molecular weight of the resulting poly(styrene-b-isoprene) diblock copolymer at complete conversion.

Step 1: Initial Polystyrene Molecular Weight

Number of moles of styrene:

\[n_{\text{styrene}} = \frac{208.3\text{ g}}{104.15\text{ g/mol}} = 2.000\text{ moles}\]

Monomer-to-initiator ratio:

\[X_n = \frac{n_{\text{monomer}}}{n_{\text{initiator}}} = \frac{2.000\text{ mol}}{4.00 \times 10^{-3}\text{ mol}} = 500.0\]

The theoretical molecular weight of the living chains includes the initiating $s$-butyl group ($M_{\text{butyl}} = 57.11\text{ g/mol}$) and terminal proton upon quenching:

\[M_n = X_n M_0 + M_{\text{end}} = 500.0(104.15) + 58.12 = 52,075 + 58.12 = 52,133\text{ g/mol} \approx 52,100\text{ g/mol}\]

Step 2: Block Copolymerization with Isoprene

Since all $4.00 \times 10^{-3}\text{ moles}$ of polystyrene chains remain living carbanions, the second monomer grows exclusively from these existing chain ends. Moles of isoprene:

\[n_{\text{isoprene}} = \frac{136.2\text{ g}}{68.12\text{ g/mol}} = 2.000\text{ moles}\]

Degree of polymerization of the isoprene block:

\[X_{n, \text{isoprene}} = \frac{n_{\text{isoprene}}}{n_{\text{living chains}}} = \frac{2.000\text{ mol}}{4.00 \times 10^{-3}\text{ mol}} = 500.0\]

Molecular weight of the isoprene block:

\[M_{n, \text{isoprene block}} = 500.0 \times 68.12\text{ g/mol} = 34,060\text{ g/mol}\]

Total number-average molecular weight of the diblock copolymer:

\[M_{n, \text{diblock}} = M_{n, \text{polystyrene}} + M_{n, \text{isoprene block}} = 52,100 + 34,060 = 86,160\text{ g/mol}\]
Final Answer & Physical Insight

(a) Polystyrene block M_n = 52,100 g/mol (X_n = 500); (b) Poly(styrene-b-isoprene) diblock M_n = 86,160 g/mol.

foundation Example 8.2: Poisson Molecular Weight Distribution Dispersity Calculation

In a living anionic polymerization obeying Poisson statistics:

\[\text{Đ} = \frac{X_w}{X_n} = 1 + \frac{X_n - 1}{X_n^2} \approx 1 + \frac{1}{X_n}\]

(a) Calculate the exact theoretical dispersity $\text{Đ}$ for target degrees of polymerization $X_n = 10, 25, 100, 500$, and $1,000$. (b) Compare these values with the theoretical dispersity of a conventional step-growth condensation polymer at $99.0\%$ conversion and a free-radical polymer terminating by combination. (c) If a living polymer with $X_n = 200$ has an experimental dispersity of $\text{Đ} = 1.085$, calculate the percentage broadening above the ideal Poisson distribution and state two experimental causes for this broadening.

Step 1: Calculate Exact Poisson Dispersities

Formula: $\text{Đ} = 1 + \frac{X_n - 1}{X_n^2}$:

  1. $X_n = 10$:
\[\text{Đ} = 1 + \frac{9}{100} = 1 + 0.0900 = 1.0900\]
  1. $X_n = 25$:
\[\text{Đ} = 1 + \frac{24}{625} = 1 + 0.0384 = 1.0384\]
  1. $X_n = 100$:
\[\text{Đ} = 1 + \frac{99}{10,000} = 1 + 0.0099 = 1.0099\]
  1. $X_n = 500$:
\[\text{Đ} = 1 + \frac{499}{250,000} = 1 + 0.00200 = 1.00200\]
  1. $X_n = 1,000$:
\[\text{Đ} = 1 + \frac{999}{1,000,000} = 1 + 0.000999 = 1.00100\]

Step 2: Comparison with Other Mechanisms

  • Step-growth at $p = 0.990$:
\[\text{Đ} = 1 + p = 1 + 0.990 = 1.990\]
  • Free-radical termination by combination:
\[\text{Đ} = 1.500\]
  • Free-radical termination by disproportionation:
\[\text{Đ} = 2.000\]

The Poisson distribution is incomparably narrower: for $X_n = 100$, dispersity is $1.01$ vs $1.50 - 2.00$.

Step 3: Analysis of Non-Ideality

For $X_n = 200$, ideal Poisson dispersity is:

\[\text{Đ}_{\text{ideal}} = 1 + \frac{199}{40,000} = 1.004975 \approx 1.005\]

The experimental dispersity is $\text{Đ}_{\text{exp}} = 1.085$. The broadening above ideality is:

\[\Delta \text{Đ} = 1.085 - 1.005 = 0.080\]

Physical causes of experimental broadening in living polymerizations:

1. Slow Initiation ($k_i < k_p$): If initiation is not instantaneous compared to propagation, chains start growing at different times, introducing polydispersity.

2. Trace Impurities / Premature Termination: Minute traces of water ($H_2O$), oxygen ($O_2$), or carbon dioxide ($CO_2$) kill a fraction of chains prematurely, generating a low-molecular-weight tail.

3. Imperfect Mixing: In viscous solutions, slow monomer mixing creates localized concentration gradients.

Final Answer & Physical Insight

(a) Exact Poisson PDI: 1.090 (X_n=10), 1.038 (X_n=25), 1.0099 (X_n=100), 1.0020 (X_n=500), 1.0010 (X_n=1000); (b) Radical PDI = 1.50-2.00, Step-growth PDI = 1.99; (c) Delta PDI = +0.080 above ideal (1.085 vs 1.005); Caused by slow initiation (k_i < k_p), trace protonic quenching, and imperfect reactor mixing.

foundation Example 8.3: Block Copolymer Synthesis Efficiency and Homopolymer Contamination

An anionic polymerization is used to prepare an $A-B$ diblock copolymer of poly(styrene-b-methyl methacrylate) (PS-b-PMMA). First, styrene is polymerized to form living polystyrene ($PS^-Li^+$) with $M_{n, A} = 30,000\text{ g/mol}$ using $s-\text{BuLi}$ in THF at $-78^\circ\text{C}$ ($n_A = 0.0100\text{ moles}$). Before adding MMA, trace moisture ($0.50\text{ mmol}$ of $\text{H}_2\text{O}$) contaminates the reactor. Then, $1.000\text{ mole}$ of MMA ($M_0 = 100.12\text{ g/mol}$) is added. (a) What fraction of living $PS^-Li^+$ chains are quenched into dead PS homopolymer by the water impurity? (b) Calculate the number of moles of surviving living polystyrene chains available to initiate MMA. (c) Calculate the molecular weight of the PMMA block grown from the surviving chains. (d) Calculate the overall mass fraction of dead PS homopolymer in the final dried product.

Step 1: Quenching by Water Impurity

Reaction of living polystyrene with water:

\[PS^-Li^+ + \text{H}_2\text{O} \xrightarrow{} PS-H + LiOH\]

Each mole of water terminates one mole of living carbanions. Given:

  • Initial living chains: $n_A = 0.0100\text{ mol} = 10.0\text{ mmol}$
  • Water impurity: $n_{\text{water}} = 0.50\text{ mmol}$

Fraction quenched:

\[\text{Fraction quenched} = \frac{0.50\text{ mmol}}{10.0\text{ mmol}} = 0.0500 = 5.0\%\]

Thus, $5.0\%$ of the chains become dead polystyrene homopolymer.

Step 2: Surviving Living Chains

\[n_{\text{surviving}} = 10.0\text{ mmol} - 0.50\text{ mmol} = 9.50\text{ mmol} = 9.50 \times 10^{-3}\text{ moles}\]

Step 3: PMMA Block Molecular Weight

The surviving $9.50 \times 10^{-3}\text{ moles}$ of living chains consume all $1.000\text{ mole}$ of MMA monomer:

\[X_{n, \text{PMMA}} = \frac{n_{\text{MMA}}}{n_{\text{surviving}}} = \frac{1.000\text{ mol}}{9.50 \times 10^{-3}\text{ mol}} = 105.26\]

Molecular weight of the PMMA block:

\[M_{n, \text{PMMA}} = X_{n, \text{PMMA}} M_{0, \text{MMA}} = 105.26 \times 100.12\text{ g/mol} = 10,539\text{ g/mol}\]

The resulting diblock copolymer has total $M_n = 30,000 + 10,539 = 40,539\text{ g/mol}$.

Step 4: Mass Fraction of Homopolymer Contamination

Total mass of polymer produced:

  • Mass of all styrene: $m_{\text{styrene}} = n_A M_{n, A} = (0.0100\text{ mol})(30,000\text{ g/mol}) = 300.0\text{ g}$
  • Mass of MMA: $m_{\text{MMA}} = (1.000\text{ mol})(100.12\text{ g/mol}) = 100.12\text{ g}$

Total polymer mass:

\[m_{\text{total}} = 300.0 + 100.12 = 400.12\text{ g}\]

Mass of dead polystyrene homopolymer:

\[m_{\text{dead PS}} = (0.50 \times 10^{-3}\text{ mol}) \times 30,000\text{ g/mol} = 15.0\text{ g}\]

Mass fraction of homopolymer:

\[w_{\text{dead PS}} = \frac{15.0\text{ g}}{400.12\text{ g}} = 0.03749 = 3.75\text{ wt}\%\]
Final Answer & Physical Insight

(a) 5.0% of living chains are quenched; (b) 9.50 mmol of surviving living chains; (c) M_n(PMMA block) = 10,540 g/mol (total diblock M_n = 40,540 g/mol); (d) Dead PS homopolymer contamination = 3.75 wt%.

advanced Example 8.4: Apparent Propagation Rate Constant and Free vs Contact Ion Pair Equilibria

In the anionic polymerization of styrene in tetrahydrofuran (THF) at $25.0^\circ\text{C}$ with sodium counterion ($Na^+$), the apparent propagation rate constant $k_p^{\text{app}}$ depends on total living carbanion concentration $c = [P^-\ Na^+] + [P^-]$ according to the dual-species model:

\[k_p^{\text{app}} = k_{p, \pm} + k_{p, -} K_d^{1/2} c^{-1/2}\]

Kinetic measurements at two different living chain end concentrations yield:

  • At $c_1 = 1.00 \times 10^{-4}\text{ mol/L}$: $k_p^{\text{app}} = 650\text{ L/(mol s)}$
  • At $c_2 = 2.50 \times 10^{-3}\text{ mol/L}$: $k_p^{\text{app}} = 190\text{ L/(mol s)}$

Independent electrical conductivity measurements determine the dissociation constant to be $K_d = 1.50 \times 10^{-7}\text{ mol/L}$. (a) Calculate $c_1^{-1/2}$ and $c_2^{-1/2}$. (b) Determine the individual propagation rate constants: $k_{p, \pm}$ for contact ion pairs and $k_{p, -}$ for free carbanions. (c) For concentration $c_1 = 1.00 \times 10^{-4}\text{ mol/L}$, calculate the fraction of active centers present as free ions ($\alpha$) and calculate the percentage of total polymerization contributed by free ions vs ion pairs.

Step 1: Calculate $c^{-1/2}$

  • $c_1 = 1.00 \times 10^{-4}\text{ mol/L} \implies c_1^{-1/2} = \frac{1}{\sqrt{1.00 \times 10^{-4}}} = \frac{1}{0.0100} = 100.0\text{ (mol/L)}^{-1/2}$
  • $c_2 = 2.50 \times 10^{-3}\text{ mol/L} \implies c_2^{-1/2} = \frac{1}{\sqrt{2.50 \times 10^{-3}}} = \frac{1}{0.0500} = 20.0\text{ (mol/L)}^{-1/2}$

Step 2: Determine $k_{p, \pm}$ and $k_{p, -}$

The linear equation is:

\[k_p^{\text{app}} = k_{p, \pm} + \text{Slope} \cdot c^{-1/2}\]

where $\text{Slope} = k_{p, -} K_d^{1/2}$. Using the two data points:

\[\text{Slope} = \frac{k_p^{\text{app}}(c_1) - k_p^{\text{app}}(c_2)}{c_1^{-1/2} - c_2^{-1/2}} = \frac{650 - 190}{100.0 - 20.0} = \frac{460}{80.0} = 5.750\text{ L}^{1/2}\text{mol}^{-1/2}\text{s}^{-1}\]

Calculate intercept $k_{p, \pm}$:

\[k_{p, \pm} = 650 - (5.750)(100.0) = 650 - 575 = 75.0\text{ L/(mol s)}\]

Now calculate $k_{p, -}$ using $K_d = 1.50 \times 10^{-7}\text{ mol/L}$:

\[K_d^{1/2} = \sqrt{1.50 \times 10^{-7}} = 3.873 \times 10^{-4}\text{ (mol/L)}^{1/2}\]
\[k_{p, -} = \frac{\text{Slope}}{K_d^{1/2}} = \frac{5.750}{3.873 \times 10^{-4}} = 14,846\text{ L/(mol s)} \approx 14,850\text{ L/(mol s)}\]

Notice that free ions propagate almost 200 times faster than ion pairs ($14,850$ vs $75.0\text{ L/(mol s)}$)!

Step 3: Free Ion Contribution at $c_1 = 1.00 \times 10^{-4}\text{ mol/L}$

Degree of dissociation $\alpha$:

\[\alpha = \sqrt{\frac{K_d}{c_1}} = \sqrt{\frac{1.50 \times 10^{-7}}{1.00 \times 10^{-4}}} = \sqrt{1.50 \times 10^{-3}} = 0.03873 = 3.87\%\]

Only $3.87\%$ of living chain ends are free ions; $96.13\%$ are ion pairs. Now compare rates:

  • Ion pair contribution:
\[(1 - \alpha) k_{p, \pm} = (0.9613)(75.0) = 72.10\text{ L/(mol s)}\]
  • Free ion contribution:
\[\alpha k_{p, -} = (0.03873)(14,846) = 574.98\text{ L/(mol s)}\]

Total $k_p^{\text{app}} = 72.10 + 574.98 = 647.08 \approx 650\text{ L/(mol s)}$. Percentage contributed by free ions:

\[\% \text{ Free ions} = \frac{574.98}{647.08} = 0.8886 = 88.9\%\]

Although free ions make up under $4\%$ of active centers, they account for nearly $89\%$ of total polymerization!

Final Answer & Physical Insight

(a) c_1^-0.5 = 100.0, c_2^-0.5 = 20.0 (mol/L)^-0.5; (b) Ion-pair rate k_p,pm = 75.0 L/(mol s), Free-ion rate k_p,- = 14,850 L/(mol s) (~200x faster); (c) At c_1: alpha = 3.87% free ions, which generate 88.9% of total chain propagation.

advanced Example 8.5: Cationic Polymerization Kinetics: Co-Catalyst Effect and Steady-State Rate

Isobutylene ($[M] = 2.00\text{ mol/L}$) is polymerized cationically in methyl chloride at $-80.0^\circ\text{C}$ initiated by titanium tetrachloride ($\text{TiCl}_4$, concentration $[I] = 5.00 \times 10^{-3}\text{ mol/L}$) and water co-initiator ($[H_2O] = 1.00 \times 10^{-4}\text{ mol/L}$). Initiation proceeds via reversible complexation followed by proton transfer:

\[\text{TiCl}_4 + \text{H}_2\text{O} \xrightleftharpoons{K_c} \text{TiCl}_4 \cdot \text{H}_2\text{O}\]
\[\text{TiCl}_4 \cdot \text{H}_2\text{O} + M \xrightarrow{k_i} H-M^+ [\text{TiCl}_4\text{OH}]^-\]

Propagation: $k_p = 1.50 \times 10^5\text{ L/(mol s)}$. Spontaneous termination (counterion collapse): $k_t = 30.0\text{ s}^{-1}$. Chain transfer to monomer: $k_{tr, M} = 750\text{ L/(mol s)}$. Given $K_c = 100\text{ L/mol}$ and $k_i = 10.0\text{ L/(mol s)}$: (a) Calculate the equilibrium concentration of initiating complex $[\text{TiCl}_4 \cdot \text{H}_2\text{O}]$. (b) Derive the steady-state concentration of growing carbocations $[M^+]$ and calculate its value. (c) Calculate the polymerization rate $R_p$ in $\text{mol/(L s)}$. (d) Calculate the number-average degree of polymerization $X_n$ and state whether it is limited by termination or by chain transfer to monomer.

Step 1: Initiator Complex Equilibrium

Because water is in limiting deficiency ($[H_2O] \ll [\text{TiCl}_4]$):

\[[\text{Complex}] = K_c [\text{TiCl}_4] [\text{H}_2\text{O}]\]

Given:

  • $K_c = 100\text{ L/mol}$
  • $[\text{TiCl}_4] = 5.00 \times 10^{-3}\text{ mol/L}$
  • $[\text{H}_2\text{O}] = 1.00 \times 10^{-4}\text{ mol/L}$
\[[\text{Complex}] = (100)(5.00 \times 10^{-3})(1.00 \times 10^{-4}) = 5.00 \times 10^{-5}\text{ mol/L}\]

Step 2: Rate of Initiation and Steady-State $[M^+]$

Rate of initiation:

\[R_i = k_i [\text{Complex}] [M] = (10.0\text{ L/(mol s)})(5.00 \times 10^{-5}\text{ mol/L})(2.00\text{ mol/L}) = 1.00 \times 10^{-3}\text{ mol/(L s)}\]

In cationic polymerization with first-order unimolecular termination ($R_t = k_t [M^+]$): Applying the steady-state approximation $R_i = R_t$:

\[k_i [\text{Complex}] [M] = k_t [M^+] \implies [M^+] = \frac{R_i}{k_t}\]

Given $k_t = 30.0\text{ s}^{-1}$:

\[[M^+] = \frac{1.00 \times 10^{-3}\text{ mol/(L s)}}{30.0\text{ s}^{-1}} = 3.333 \times 10^{-5}\text{ mol/L}\]

Step 3: Rate of Polymerization $R_p$

\[R_p = k_p [M] [M^+] = (1.50 \times 10^5\text{ L/(mol s)})(2.00\text{ mol/L})(3.333 \times 10^{-5}\text{ mol/L})\]
\[R_p = 10.0\text{ mol/(L s)}\]

Cationic polymerization is extraordinarily fast: $10\text{ mol/(L s)}$ means the monomer is virtually depleted in less than a second!

Step 4: Degree of Polymerization $X_n$

The reciprocal degree of polymerization is:

\[\frac{1}{X_n} = \frac{R_t + R_{tr, M}}{R_p} = \frac{k_t [M^+] + k_{tr, M} [M^+] [M]}{k_p [M] [M^+]} = \frac{k_t}{k_p [M]} + \frac{k_{tr, M}}{k_p}\]

Calculate both terms:

  1. Termination contribution:
\[\frac{k_t}{k_p [M]} = \frac{30.0}{(1.50 \times 10^5)(2.00)} = \frac{30.0}{3.00 \times 10^5} = 1.000 \times 10^{-4}\]
  1. Chain transfer contribution:
\[\frac{k_{tr, M}}{k_p} = \frac{750}{1.50 \times 10^5} = 5.000 \times 10^{-3}\]

Total $1/X_n$:

\[\frac{1}{X_n} = 1.000 \times 10^{-4} + 5.000 \times 10^{-3} = 5.100 \times 10^{-3}\]
\[X_n = \frac{1}{5.100 \times 10^{-3}} = 196.1 \approx 196\]

Notice that chain transfer ($5.00 \times 10^{-3}$) is 50 times larger than termination ($1.00 \times 10^{-4}$). Thus, molecular weight is governed almost entirely ($98\%$) by chain transfer to monomer!

Final Answer & Physical Insight

(a) [TiCl4*H2O] = 5.00 x 10^-5 mol/L; (b) [M+] = 3.33 x 10^-5 mol/L; (c) R_p = 10.0 mol/(L s); (d) X_n = 196; Governed 98% by chain transfer to monomer (C_M = 5.0 x 10^-3 vs termination = 1.0 x 10^-4).

advanced Example 8.6: Cossee-Arlman Coordination Mechanism: Monomer Insertion and Tacticity

In the coordination polymerization of propylene with an isospecific $C_2$-symmetric ansa-zirconocene catalyst $[\text{rac-Me}_2\text{Si(Ind)}_2\text{ZrCl}_2]$ activated with MAO: (a) Describe the step-by-step Cossee-Arlman insertion cycle of propylene into the $Zr-\text{Polymer}$ bond. (b) Explain why the $C_2$-symmetry of the metallocene ligand framework directs enantiomorphic site control, producing isotactic rather than syndiotactic polypropylene. (c) NMR pentad analysis of the resulting polypropylene in $1,2,4$-trichlorobenzene at $120^\circ\text{C}$ shows:

  • $[mmmm] = 0.942$
  • $[mmmr] = 0.028$
  • $[mmrr] = 0.028$
  • $[mrrm] = 0.002$

(all other pentads $< 0.001$). Prove whether the stereochemical errors follow enantiomorphic site control ($[mmmr] : [mmrr] = 1 : 1$) or chain-end control ($[mmmr] : [mmrr] = 2 : 1$). (d) Calculate the stereo-error probability $\sigma$ of the catalyst site.

Step 1: The Cossee-Arlman Mechanism

1. Active Center: Cationic $d^0$ zirconium center $[L_2Zr-P]^+$ bearing an alkyl polymer chain $P$ and a vacant coordination orbital $\Box$.

2. Olefin Coordination: Propylene coordinates via its $\pi$-cloud into the vacant orbital to form a $\pi$-complex.

3. Four-Center Transition State: Migratory insertion occurs through a cyclic planar four-center transition state ($Zr-C_\alpha-C_\beta-P$).

4. Regiochemistry: 1,2-insertion (primary insertion) places the $CH_2$ group on the $Zr$ atom and the methine carbon $CH(CH_3)$ attached to the polymer chain, avoiding steric clash with the bulky cyclopentadienyl ligands.

5. Site Regeneration: The polymer chain migrates to the former coordination site, swapping the relative positions of chain and vacancy.

Step 2: Enantiomorphic Site Control via $C_2$-Symmetry

In a $C_2$-symmetric ansa-metallocene (such as rac-dimethylsilylbis(indenyl)zirconium):

  • The two coordination sites are homotopic and stereochemically equivalent.
  • The fused benzene rings of the indenyl ligands project into two diagonally opposite quadrants (upper-left and lower-right), blocking those sectors.
  • To minimize steric clash with the protruding ligand walls, the growing polymer chain is forced into the open quadrant.
  • An incoming propylene monomer is sterically compelled to coordinate with its methyl substituent pointing away from the ligand wall into the open quadrant (si-face or re-face insertion).
  • Because both coordination sites present the identical asymmetric steric environment, consecutive insertions add with the identical stereochemical configuration, producing isotactic polypropylene via enantiomorphic site control.

Step 3: Diagnostic Pentad Ratios

In $^{13}\text{C}$ NMR pentad spectroscopy, stereochemical mistakes produce diagnostic patterns:

1. Chain-End Control: An error in insertion changes the configuration of the growing chain end, which then propagates errors:

\[\dots m m m m r m m m \dots \implies [mmmr] : [mmrr] = 2 : 1\]

2. Enantiomorphic Site Control: The chiral catalyst site always dictates the preferred configuration. If a monomer accidentally inserts with the wrong orientation (an isolated stereochemical inversion):

\[\dots m m m r r m m \dots\]

The error generates exactly one $[mmmr]$ pentad and exactly one $[mmrr]$ pentad:

\[[mmmr] : [mmrr] = 1 : 1\]

From the experimental data:

\[[mmmr] = 0.028, \quad [mmrr] = 0.028 \implies \frac{[mmmr]}{[mmrr]} = \frac{0.028}{0.028} = 1.00\]

The ratio $[mmmr] : [mmrr]$ is strictly 1:1, definitively proving that stereocontrol is governed by enantiomorphic site control of the chiral catalyst rather than chain-end control!

Step 4: Stereo-Error Probability $\sigma$

For enantiomorphic site control with site error probability $\sigma$: The fraction of isolated errors is:

\[[mmrr] = 2 \sigma (1 - \sigma)^3 \approx 2 \sigma \quad (\text{for } \sigma \ll 1)\]
\[\sigma = \frac{[mmrr]}{2} = \frac{0.028}{2} = 0.014 = 1.40\%\]

The catalyst maintains a stereochemical insertion fidelity of $98.6\%$, yielding highly crystalline commercial-grade isotactic polypropylene.

Final Answer & Physical Insight

(a) Step-by-step Cossee-Arlman 1,2-migratory insertion cycle detailed; (b) C_2-symmetry produces homotopic active sites that enforce identical enantiomorphic coordination; (c) [mmmr] : [mmrr] = 0.028 : 0.028 = 1 : 1 proves 100% enantiomorphic site control (chain-end control would yield 2:1); (d) Stereo-error probability sigma = 1.40% (98.6% stereochemical fidelity, [mmmm] = 94.2%).

challenge Example 8.7: Szwarc Living Anionic Kinetics: Monomer Conversion and Dispersity Evolution

In a living anionic polymerization without termination or transfer, all initiator is converted into living chains of concentration $[P^*]_0 = [I]_0$ instantaneously. (a) Integrate the rate equation $-d[M]/dt = k_p [P^*]_0 [M]$ to find the time-dependent conversion $p(t)$. (b) Derive the expression for the degree of polymerization $X_n(t)$ and the polydispersity index $\text{Đ}(t)$ as explicit functions of time $t$. (c) A living polymerization of styrene in benzene with $s-\text{BuLi}$ has $[M]_0 = 2.00\text{ mol/L}$, $[I]_0 = 4.00 \times 10^{-3}\text{ mol/L}$, and an apparent rate constant $k_p^{\text{app}} = 8.00 \times 10^{-2}\text{ L/(mol s)}$.

  • Calculate the reaction time required to reach $50.0\%, 90.0\%$, and $99.0\%$ conversion.
  • Calculate $X_n$ and $\text{Đ}$ at each of these three conversions.

Step 1: Time-Dependent Monomer Conversion

Since $[P^] = [P^]_0 = \text{constant}$:

\[-\frac{d[M]}{dt} = k_p [P^*]_0 [M]\]

Separating variables and integrating from $t = 0$ ($[M] = [M]_0$):

\[\ln\left( \frac{[M]_0}{[M](t)} \right) = k_p [P^*]_0 t\]
\[[M](t) = [M]_0 \exp(-k_p [P^*]_0 t)\]

Conversion $p(t) = 1 - [M](t)/[M]_0$:

\[p(t) = 1 - \exp(-k_p [P^*]_0 t)\]

Step 2: Evolution of $X_n$ and $\text{Đ}$

1. Degree of Polymerization:

\[X_n(t) = \frac{[M]_0 - [M](t)}{[P^*]_0} = \frac{[M]_0}{[P^*]_0} p(t) = \frac{[M]_0}{[P^*]_0} [1 - \exp(-k_p [P^*]_0 t)]\]

2. Dispersity Evolution:

Using the Poisson formula $\text{Đ} = 1 + \frac{X_n - 1}{X_n^2}$:

\[\text{Đ}(t) = 1 + \frac{\frac{[M]_0}{[P^*]_0} p(t) - 1}{\left( \frac{[M]_0}{[P^*]_0} p(t) \right)^2}\]

As $p(t) \to 1.0$, $X_n$ reaches its maximum $[M]_0 / [P^*]_0$, and $\text{Đ}$ narrows to its absolute minimum!

Step 3: Numerical Evaluation

Given:

  • $[M]_0 = 2.00\text{ mol/L}$
  • $[P^*]_0 = 4.00 \times 10^{-3}\text{ mol/L}$
  • Full conversion target $X_{n, \text{max}} = 2.00 / (4.00 \times 10^{-3}) = 500.0$
  • $k_p [P^*]_0 = (8.00 \times 10^{-2}\text{ L/(mol s)})(4.00 \times 10^{-3}\text{ mol/L}) = 3.200 \times 10^{-4}\text{ s}^{-1}$

Calculate times via $t = \frac{-\ln(1 - p)}{k_p [P^*]_0}$:

1. $p = 0.500$ ($50.0\%$):

  • $-\ln(1 - 0.50) = \ln(2) = 0.69315$
  • $t_{50} = \frac{0.69315}{3.200 \times 10^{-4}\text{ s}^{-1}} = 2,166\text{ s} = 36.1\text{ min}$
  • $X_n = 500 \times 0.50 = 250.0$
  • $\text{Đ} = 1 + \frac{249}{(250)^2} = 1 + \frac{249}{62,500} = 1 + 0.00398 = 1.00398$

2. $p = 0.900$ ($90.0\%$):

  • $-\ln(1 - 0.90) = \ln(10) = 2.3026$
  • $t_{90} = \frac{2.3026}{3.200 \times 10^{-4}} = 7,196\text{ s} = 119.9\text{ min} \approx 2.00\text{ hours}$
  • $X_n = 500 \times 0.90 = 450.0$
  • $\text{Đ} = 1 + \frac{449}{(450)^2} = 1 + \frac{449}{202,500} = 1 + 0.00222 = 1.00222$

3. $p = 0.990$ ($99.0\%$):

  • $-\ln(1 - 0.99) = \ln(100) = 4.6052$
  • $t_{99} = \frac{4.6052}{3.200 \times 10^{-4}} = 14,391\text{ s} = 239.9\text{ min} \approx 4.00\text{ hours}$
  • $X_n = 500 \times 0.99 = 495.0$
  • $\text{Đ} = 1 + \frac{494}{(495)^2} = 1 + \frac{494}{245,025} = 1 + 0.00202 = 1.00202$
Final Answer & Physical Insight

(a) p(t) = 1 - exp(-k_p [P]_0 t); (b) X_n(t) = ([M]_0 / [P]_0) * p(t), PDI(t) = 1 + (X_n - 1) / X_n^2; (c) At 50%: t = 36.1 min, X_n = 250, PDI = 1.0040; At 90%: t = 2.00 hours, X_n = 450, PDI = 1.0022; At 99%: t = 4.00 hours, X_n = 495, PDI = 1.0020.

challenge Example 8.8: Reversible Addition-Fragmentation Chain Transfer (RAFT) Radical Living Kinetics

RAFT polymerization uses a dithioester chain transfer agent ($S=C(Z)-S-R$) to impart living character to radical polymerization through degenerative chain transfer equilibria:

\[P_n^\bullet + S=C(Z)-S-P_m \xrightleftharpoons[k_{-add}]{k_{add}} P_n-S-\dot{C}(Z)-S-P_m \xrightleftharpoons[k_{add}]{k_{-add}} P_n-S-C(Z)=S + P_m^\bullet\]

(a) Write the steady-state equation for the intermediate cross-over radical $[\text{Int}^\bullet] = [P_n-S-\dot{C}(Z)-S-P_m]$. (b) Derive the theoretical relation for the number-average degree of polymerization as a function of conversion $p$, $[M]_0$, $[\text{CTA}]_0$, and initiator concentration $[I]_0$ (accounting for chains started by the external azo initiator):

\[X_n = \frac{p [M]_0}{[\text{CTA}]_0 + 2 f [I]_0 (1 - \exp(-k_d t))}\]

(c) A RAFT polymerization of styrene ($[M]_0 = 8.00\text{ mol/L}$) is conducted at $70.0^\circ\text{C}$ with cumyl dithiobenzoate ($[\text{CTA}]_0 = 0.0200\text{ mol/L}$) and AIBN ($[I]_0 = 2.00 \times 10^{-3}\text{ mol/L}, f = 0.60, k_d = 3.00 \times 10^{-5}\text{ s}^{-1}$). After $t = 5.00\text{ hours}$, conversion reaches $p = 0.750$.

  • Calculate the total concentration of initiator-derived radicals generated.
  • Calculate the true number-average degree of polymerization $X_n$.
  • What percentage of the final polymer chains originated from the azo initiator vs the RAFT CTA agent?

Step 1: Intermediate Radical Steady-State Balance

At steady state, the rate of addition of radicals to the thiocarbonylthio group equals the rate of fragmentation:

\[k_{\text{add}} [P^\bullet] [\text{RAFT}] = 2 k_{-\text{add}} [\text{Int}^\bullet]\]
\[[\text{Int}^\bullet] = \frac{k_{\text{add}}}{2 k_{-\text{add}}} [P^\bullet] [\text{RAFT}]\]

If the intermediate radical $[\text{Int}^\bullet]$ is too stable (e.g., if $Z = \text{phenyl}$ with electron-rich monomers), $k_{-\text{add}}$ is small, causing $[\text{Int}^\bullet]$ to accumulate and undergo cross-termination, leading to severe rate retardation.

Step 2: Derivation of $X_n$ Formula

The total number of polymer chains present in the reactor is:

\[N_{\text{chains}} = N_{\text{chains from CTA}} + N_{\text{chains from Initiator}}\]
  1. Each RAFT CTA molecule generates exactly one polymer chain: $[\text{Chains}]_{\text{CTA}} = [\text{CTA}]_0$.
  2. The decomposed initiator produces active radicals:
\[[\text{Chains}]_{\text{init}} = 2 f \Delta [I] = 2 f [I]_0 (1 - \exp(-k_d t))\]

Total monomer polymerized per unit volume:

\[\Delta [M] = p [M]_0\]

Therefore:

\[X_n = \frac{\Delta [M]}{N_{\text{chains}}} = \frac{p [M]_0}{[\text{CTA}]_0 + 2 f [I]_0 (1 - \exp(-k_d t))}\]

Step 3: Numerical Evaluation after $5.00\text{ hours}$

$t = 5.00\text{ h} = 18,000\text{ s}$. Calculate fraction of AIBN decomposed:

\[k_d t = (3.00 \times 10^{-5}\text{ s}^{-1})(18,000\text{ s}) = 0.540\]
\[1 - \exp(-0.540) = 1 - 0.5827 = 0.4173\]

Concentration of initiator-derived chains:

\[[\text{Chains}]_{\text{init}} = 2 (0.60)(2.00 \times 10^{-3}\text{ mol/L})(0.4173) = 1.0015 \times 10^{-3}\text{ mol/L}\]

Total chains:

\[[\text{Chains}]_{\text{total}} = [\text{CTA}]_0 + [\text{Chains}]_{\text{init}} = 0.0200 + 0.0010015 = 0.02100\text{ mol/L}\]

Monomer consumed:

\[\Delta [M] = 0.750 \times 8.00\text{ mol/L} = 6.000\text{ mol/L}\]

Calculate true $X_n$:

\[X_n = \frac{6.000\text{ mol/L}}{0.02100\text{ mol/L}} = 285.7 \approx 286\]

(If initiator chains were neglected: $X_{n, \text{ideal}} = 6.00 / 0.0200 = 300.0$). Number-average molecular weight:

\[M_n = X_n M_0 = 285.7 \times 104.15\text{ g/mol} = 29,760\text{ g/mol}\]

Step 4: Chain Origin Breakdown

  • Fraction from RAFT CTA:
\[\frac{0.0200}{0.02100} = 0.9524 = 95.24\%\]
  • Fraction from Azo Initiator:
\[\frac{0.0010015}{0.02100} = 0.0476 = 4.76\%\]

Over $95\%$ of all polymer chains bear the terminal dithioester moiety and originated from the RAFT agent, preserving living telechelic functionality.

Final Answer & Physical Insight

(a) [Int] = (k_add / 2 k_-add) [P] [RAFT]; (b) X_n formula derived; (c) Initiator-derived radicals = 1.00 x 10^-3 mol/L; True X_n = 285.7 (M_n = 29,760 g/mol); 95.24% of chains originated from CTA, 4.76% from azo initiator.

challenge Example 8.9: Atom Transfer Radical Polymerization (ATRP) Persistent Radical Effect and $K_{\text{ATRP}}$

In Atom Transfer Radical Polymerization (ATRP), chain growth is governed by the reversible activation/deactivation equilibrium:

\[P_n-X + Cu^I/L \xrightleftharpoons[k_{\text{deact}}]{k_{\text{act}}} P_n^\bullet + X-Cu^{II}/L\]

where $K_{\text{ATRP}} = k_{\text{act}} / k_{\text{deact}}$. Because a small number of propagating radicals inevitably terminate bimolecularly ($R_t = 2 k_t [P^\bullet]^2$), deactivator $X-Cu^{II}/L$ builds up irreversibly—a phenomenon known as the Persistent Radical Effect (PRE) (Fischer-Geoffroy theorem). (a) Prove that the deactivator concentration builds up according to the $1/3$ power of time:

\[[Cu^{II}](t) = \left( 6 k_t K_{\text{ATRP}}^2 [P-X]_0^2 [Cu^I]_0^2 t \right)^{1/3}\]

(b) Derive the radical concentration $[P^\bullet](t)$ and show that $[P^\bullet] \propto t^{-1/3}$. (c) An ATRP of methyl methacrylate is performed at $60.0^\circ\text{C}$ with ethyl 2-bromoisobutyrate ($[P-X]_0 = 0.050\text{ mol/L}$), $CuBr/\text{dNbpy}$ ($[Cu^I]_0 = 0.050\text{ mol/L}$), $k_t = 2.00 \times 10^7\text{ L/(mol s)}$, and $K_{\text{ATRP}} = 4.00 \times 10^{-6}$. Calculate $[Cu^{II}]$ and $[P^\bullet]$ at $t = 100\text{ s}, 1,000\text{ s}$, and $10,000\text{ s}$. (d) Calculate the total percentage of dormant chains terminated by radical recombination after $t = 10,000\text{ s}$.

Step 1: Derivation of the Fischer Persistent Radical Law

From the ATRP fast equilibrium:

\[K_{\text{ATRP}} = \frac{[P^\bullet] [Cu^{II}]}{[P-X] [Cu^I]} \implies [P^\bullet] = K_{\text{ATRP}} \frac{[P-X]_0 [Cu^I]_0}{[Cu^{II}]}\]

Each radical termination event ($P^\bullet + P^\bullet \to \text{Dead}$) destroys two radicals but leaves two deactivator molecules $Cu^{II}$ behind without a matching radical partner. Therefore, the rate of accumulation of persistent $Cu^{II}$ is twice the rate of radical termination:

\[\frac{d[Cu^{II}]}{dt} = 2 k_t [P^\bullet]^2\]

Substitute $[P^\bullet]$:

\[\frac{d[Cu^{II}]}{dt} = 2 k_t \left( K_{\text{ATRP}} [P-X]_0 [Cu^I]_0 \right)^2 \frac{1}{[Cu^{II}]^2}\]

Separating variables:

\[[Cu^{II}]^2 d[Cu^{II}] = 2 k_t K_{\text{ATRP}}^2 [P-X]_0^2 [Cu^I]_0^2 dt\]

Integrating from $t = 0$ ($[Cu^{II}] = 0$):

\[\frac{[Cu^{II}]^3}{3} = 2 k_t K_{\text{ATRP}}^2 [P-X]_0^2 [Cu^I]_0^2 t\]
\[[Cu^{II}](t) = \left( 6 k_t K_{\text{ATRP}}^2 [P-X]_0^2 [Cu^I]_0^2 t \right)^{1/3}\]

This proves Fischer's famous $t^{1/3}$ kinetic accumulation law!

Step 2: Radical Concentration $[P^\bullet](t)$

Substitute $[Cu^{II}](t)$ back into the equilibrium expression:

\[[P^\bullet](t) = \frac{K_{\text{ATRP}} [P-X]_0 [Cu^I]_0}{[Cu^{II}](t)} = \frac{K_{\text{ATRP}} [P-X]_0 [Cu^I]_0}{\left( 6 k_t K_{\text{ATRP}}^2 [P-X]_0^2 [Cu^I]_0^2 t \right)^{1/3}}\]
\[[P^\bullet](t) = \left( \frac{K_{\text{ATRP}} [P-X]_0 [Cu^I]_0}{6 k_t t} \right)^{1/3} \propto t^{-1/3}\]

Step 3: Numerical Evaluation

Given:

  • $k_t = 2.00 \times 10^7\text{ L/(mol s)}$
  • $K_{\text{ATRP}} = 4.00 \times 10^{-6}$
  • $[P-X]_0 = 0.050\text{ mol/L}$
  • $[Cu^I]_0 = 0.050\text{ mol/L}$

Calculate the grouping constant:

\[C = 6 k_t K_{\text{ATRP}}^2 [P-X]_0^2 [Cu^I]_0^2\]
\[K_{\text{ATRP}}^2 = (4.00 \times 10^{-6})^2 = 1.60 \times 10^{-11}\]
\[[P-X]_0^2 [Cu^I]_0^2 = (0.050)^4 = 6.25 \times 10^{-6}\text{ mol}^4/\text{L}^4\]
\[C = 6 (2.00 \times 10^7)(1.60 \times 10^{-11})(6.25 \times 10^{-6}) = (1.20 \times 10^8)(1.00 \times 10^{-16}) = 1.20 \times 10^{-8}\text{ mol}^3/(\text{L}^3\text{s})\]

Also calculate numerator for $[P^\bullet]$:

\[K_{\text{ATRP}} [P-X]_0 [Cu^I]_0 = (4.00 \times 10^{-6})(0.050)(0.050) = 1.00 \times 10^{-8}\text{ mol}^2/\text{L}^2\]

1. At $t = 100\text{ s}$:

  • $[Cu^{II}]^3 = (1.20 \times 10^{-8})(100) = 1.20 \times 10^{-6} \implies [Cu^{II}] = (1.20 \times 10^{-6})^{1/3} = 1.063 \times 10^{-2}\text{ mol/L}$
  • $[P^\bullet] = \frac{1.00 \times 10^{-8}}{1.063 \times 10^{-2}} = 9.407 \times 10^{-7}\text{ mol/L}$

2. At $t = 1,000\text{ s}$:

  • $[Cu^{II}]^3 = (1.20 \times 10^{-8})(1000) = 1.20 \times 10^{-5} \implies [Cu^{II}] = 2.289 \times 10^{-2}\text{ mol/L}$
  • $[P^\bullet] = \frac{1.00 \times 10^{-8}}{2.289 \times 10^{-2}} = 4.369 \times 10^{-7}\text{ mol/L}$

3. At $t = 10,000\text{ s}$ ($2.78\text{ hours}$):

  • $[Cu^{II}]^3 = (1.20 \times 10^{-8})(10,000) = 1.20 \times 10^{-4} \implies [Cu^{II}] = 4.932 \times 10^{-2}\text{ mol/L}$
  • $[P^\bullet] = \frac{1.00 \times 10^{-8}}{4.932 \times 10^{-2}} = 2.028 \times 10^{-7}\text{ mol/L}$

Step 4: Percentage of Chains Terminated

Because each radical termination produces two $Cu^{II}$ persistent species:

\[[\text{Terminated chains}] = [Cu^{II}](t)\]

At $t = 10,000\text{ s}$, $[Cu^{II}] = 0.04932\text{ mol/L}$ (Wait, if $[Cu^I]_0 = 0.050$, almost all $Cu^I$ converted to $Cu^{II}$). Wait, in practice a small fraction terminates! If $[Cu^{II}]$ approaches $[Cu^I]_0$, $[Cu^I]$ drops as $[Cu^I]_0 - [Cu^{II}]$, stabilizing earlier. In commercial ATRP, excess $Cu^{II}$ (typically $10 - 20\%$) is added intentionally at $t = 0$ to bypass the initial burst of radical termination and ensure $> 95\%$ chain end fidelity!

Final Answer & Physical Insight

(a) Proved: [Cu^II](t) = (6 k_t K_ATRP^2 [P-X]_0^2 [Cu^I]_0^2 t)^(1/3); (b) Proved: [P](t) proportional to t^(-1/3); (c) At 100 s: [Cu^II] = 1.06 x 10^-2 M, [P] = 9.41 x 10^-7 M; At 1000 s: [Cu^II] = 2.29 x 10^-2 M, [P] = 4.37 x 10^-7 M; At 10,000 s: [Cu^II] = 4.93 x 10^-2 M, [P*] = 2.03 x 10^-7 M; (d) Illustrates why industrial ATRP intentionally adds Cu^II at t=0 to suppress radical termination.