University Chemistry — Year 2 · Bachelor Year 2

29Polymers: Synthesis, Structure and Properties

Nylon stockings went on sale in 1939. They came out of a laboratory that had set out, a few years earlier, to understand how small molecules join into long ones, and that had learnt along the way a lesson that surprises every chemist the first time: for a polymer made by linking molecules end to end, a 99 % yield is not enough. This chapter explains why, describes the two ways in which chains grow, and connects the structure of the chains to the properties of the materials.

You already know

The school volume: polymers, monomers and repeat units, addition and condensation polymers, thermoplastics and thermosets. The Year 1 volume: radicals and homolysis, carbocations and carbanions, the steady-state approximation. Chapter 24: esters and amides. Chapter 20: insertion into a metal–carbon bond.

29.1 Chains and their molar masses

Definition 29.1 (Polymer)

A polymer is a substance made of macromolecules, each built by the repeated linking of small molecules, the monomers. The repeat unit is the smallest group of atoms whose repetition makes up the chain; the degree of polymerisation $X$ of a chain is the number of monomer-derived units it contains.

A sample of a synthetic polymer is never made of identical chains: it is a mixture of chains of different lengths, and its molar mass is an average, which depends on how the chains are counted.

Definition 29.2 (Molar-mass averages)

For a sample containing $N_i$ chains of molar mass $M_i$ , the number-average molar mass and the mass-average molar mass are

\bar M_n = \frac{\sum_i N_i M_i}{\sum_i N_i}, \qquad \bar M_w = \frac{\sum_i N_i M_i^2}{\sum_i N_i M_i} = \sum_i w_i M_i,

where $w_i$ is the mass fraction of the chains of mass $M_i$ . The dispersity $\text{\DH} = \bar M_w/\bar M_n$ is at least 1, and equal to 1 only if all chains have the same mass. The averages $\bar X_n$ and $\bar X_w$ of the degree of polymerisation are defined in the same way.

The number average weights each chain once; the mass average weights each chain by its mass, so the long chains count more. That $\text{\DH} \ge 1$ is the inequality $\left(\sum N_i M_i\right)^2 \le \sum N_i \sum N_i M_i^2$ (Cauchy–Schwarz), with equality only for equal $M_i$ .

Method 29.3 (Computing the averages)

Turn the data into amounts $N_i$ (or numbers of chains) and masses $M_i$ ; from mass fractions, $N_i \propto w_i/M_i$ .
$\bar M_n = \sum N_i M_i / \sum N_i$ ; equivalently, $1/\bar M_n = \sum w_i/M_i$ .
$\bar M_w = \sum w_i M_i$ .
$\text{\DH} = \bar M_w/\bar M_n$ ; check that $\bar M_w \ge \bar M_n$ .

29.2 Step growth and chain growth

Definition 29.4 (Step and chain growth)

In a step-growth polymerisation, any two molecules carrying complementary functional groups (monomers, oligomers or long chains) can react, and the chains lengthen by joining one another: polyesters, polyamides. In a chain-growth polymerisation, a reactive centre (radical, ion or metal–carbon bond) adds monomer molecules one at a time to the end of a growing chain: the polymers of alkenes.

Method 29.5 (Step or chain?)

A monomer with two functional groups that react with each other’s partner (acid and alcohol, acid and amine), often releasing a small molecule: step growth.
A monomer with a $\ce{C=C}$ double bond (or a strained ring) and an initiator: chain growth.
Check with the course of the reaction: in step growth, monomer disappears early and long chains appear only at the very end; in chain growth, long chains appear from the start while monomer is consumed gradually.

Step growth

Take a balanced mixture of $\ce{A-A}$ and $\ce{B-B}$ monomers (a diamine and a diacid), or a single $\ce{A-B}$ monomer. Call $p$ the extent of reaction: the fraction of the functional groups of one kind that have reacted.

Theorem 29.6 (Carothers equation)

In a step-growth polymerisation of exactly balanced bifunctional monomers, the number-average degree of polymerisation, counted in monomer units, is

\bar X_n = \frac{1}{1 - p}.

Proof. Start with $N_0$ monomer molecules, carrying $N_0$ groups $\ce{A}$ and $N_0$ groups $\ce{B}$ . Each reaction between an $\ce{A}$ and a $\ce{B}$ joins two molecules into one, so it lowers the number of molecules by one. After $pN_0$ reactions there remain $N = N_0 - pN_0 = N_0(1-p)$ molecules sharing the $N_0$ monomer units: $\bar X_n = N_0/N = 1/(1-p)$ . ∎

Corollary 29.7 (Imbalance)

If the groups are not balanced, with $r = N_{\ce{A}}/N_{\ce{B}} \le 1$ and $p$ the extent of reaction of the minority groups $\ce{A}$ ,

\bar X_n = \frac{1 + r}{1 + r - 2rp}, \qquad \text{at most } \frac{1+r}{1-r} \text{ when } p = 1.

Proof. There are $(N_{\ce{A}} + N_{\ce{B}})/2$ bifunctional monomer molecules at the start. Each linear molecule has two chain ends, each an unreacted group; after reaction there remain $N_{\ce{A}}(1-p)$ groups $\ce{A}$ and $N_{\ce{B}} - pN_{\ce{A}}$ groups $\ce{B}$ , hence $N = [N_{\ce{A}}(1-p) + N_{\ce{B}} - pN_{\ce{A}}]/2$ molecules. Dividing, with $N_{\ce{B}} = N_{\ce{A}}/r$ : $\bar X_n = (1 + 1/r)/(1 - 2p + 1/r) = (1+r)/(1 + r - 2rp)$ . With $r = 1$ this is the Carothers equation. ∎

The equation explains the hook. At $p = 0.99$ , a “99 % yield” of the linking reaction, $\bar X_n = 100$ ; a useful fibre needs about that or more, so the linking must go beyond 99 %, with pure monomers in exactly equal amounts and the small molecule released (water, methanol) removed to keep pushing the equilibrium. An excess of 1 % of one monomer caps $\bar X_n$ at 201 even at complete conversion.

Theorem 29.8 (Flory distribution)

In a step-growth polymerisation at extent $p$ , if every group has the same reactivity whatever the length of its chain, the number fraction and the mass fraction of chains of $x$ units are

n_x = (1-p)\,p^{\,x-1}, \qquad w_x = x(1-p)^2 p^{\,x-1},

and $\bar X_n = 1/(1-p)$ , $\bar X_w = (1+p)/(1-p)$ , so that $\text{\DH} = 1 + p$ , close to 2 at high conversion.

Proof. Follow a chain from one end (an $\ce{A-B}$ monomer, for simplicity). Each link along it is formed with probability $p$ , independently. The chain has exactly $x$ units if the first $x-1$ links are formed and the next is not: probability $p^{\,x-1}(1-p)$ , which is $n_x$ . With $\sum_{x\ge1} p^{\,x-1} = 1/(1-p)$ and its derivatives $\sum x p^{\,x-1} = 1/(1-p)^2$ and $\sum x^2 p^{\,x-1} = (1+p)/(1-p)^3$ : $\sum n_x = 1$ , $\bar X_n = \sum x n_x = 1/(1-p)$ , $w_x = x n_x/\bar X_n = x(1-p)^2p^{\,x-1}$ , and $\bar X_w = \sum x w_x = (1-p)^2(1+p)/(1-p)^3 = (1+p)/(1-p)$ . ∎

Step-growth polymerisation (models). Left: Flory mass distributions (Theorem 29.8); the peak sits near

\bar X_n

and the distribution widens as it moves out. Right: the Carothers equation;

\bar X_n

stays small until

p

is very close to 1.

Radical chain growth

Definition 29.9 (Steps of a chain polymerisation)

A radical chain polymerisation runs in three kinds of steps. In initiation, a radical initiator (a molecule with a weak bond, such as a peroxide or an azo compound) splits on heating into radicals, one of which adds to a monomer. In propagation, the radical at the chain end adds a monomer molecule and remains a radical. In termination, two radicals destroy each other, by combination (they bond) or disproportionation (one takes a hydrogen atom from the other).

\begin{align*} &\text{initiation:} && \mathrm{I{-}I} \longrightarrow 2\,\mathrm{I^{\bullet}}, \qquad \mathrm{I^{\bullet} + CH_2{=}CHPh \longrightarrow I{-}CH_2{-}\dot{C}HPh} \\ &\text{propagation:} && {\sim}\mathrm{CH_2{-}\dot{C}HPh + CH_2{=}CHPh \longrightarrow}\; {\sim}\mathrm{CH_2{-}CHPh{-}CH_2{-}\dot{C}HPh} \\ &\text{termination:} && {\sim}\mathrm{CH_2{-}\dot{C}HPh + Ph\dot{C}H{-}CH_2}{\sim} \longrightarrow {\sim}\mathrm{CH_2{-}CHPh{-}CHPh{-}CH_2}{\sim} \end{align*}

Radical polymerisation of styrene. The initiator

\ce{I-I}

(dibenzoyl peroxide, for instance) splits into two radicals; each adds to the

\ce{CH2}

end of styrene, giving the more stable benzylic radical; propagation repeats the addition thousands of times; two chain radicals combine to end both chains. The dot marks the unpaired electron.

Theorem 29.10 (Rate of radical polymerisation)

With an initiator decomposing with rate constant $k_d$ and efficiency $f$ (the fraction of radicals that start a chain), propagation constant $k_p$ and termination constant $k_t$ (termination rate $2k_t [\mathrm{M^\bullet}]^2$ ), the steady-state rate of polymerisation is

R_p = k_p [\mathrm M] \sqrt{\frac{f k_d [\mathrm I]}{k_t}}.

Proof. Let $[\mathrm{M^\bullet}]$ be the total concentration of chain radicals, whatever their length. They are created at the rate $R_i = 2fk_d[\mathrm I]$ (two radicals per initiator molecule, a fraction $f$ of which start chains) and destroyed at the rate $2k_t[\mathrm{M^\bullet}]^2$ ; propagation turns one chain radical into another and does not change their number. The steady-state approximation on $[\mathrm{M^\bullet}]$ gives $2fk_d[\mathrm I] = 2k_t[\mathrm{M^\bullet}]^2$ , so $[\mathrm{M^\bullet}] = \sqrt{fk_d[\mathrm I]/k_t}$ . Monomer is consumed essentially by propagation, at $R_p = k_p[\mathrm M][\mathrm{M^\bullet}]$ . ∎

Definition 29.11 (Kinetic chain length)

The kinetic chain length $\nu$ is the average number of monomer molecules added per radical that starts a chain: $\nu = R_p/R_i$ .

Proposition 29.12 (Chain length and initiator)

In the steady state, $\nu = k_p[\mathrm M]/\bigl(2\sqrt{fk_dk_t[\mathrm I]}\bigr)$ : more initiator gives faster polymerisation but shorter chains. With termination by combination, $\bar X_n = 2\nu$ ; by disproportionation, $\bar X_n = \nu$ .

Proof. $\nu = R_p/R_i = k_p[\mathrm M]\sqrt{fk_d[\mathrm I]/k_t}/(2fk_d[\mathrm I]) = k_p[\mathrm M]/\bigl(2\sqrt{fk_dk_t[\mathrm I]}\bigr)$ . Each chain started adds $\nu$ monomers on average; combination joins two such chains into one molecule, disproportionation leaves two. ∎

Definition 29.13 (Living polymerisation)

A living polymerisation is a chain-growth polymerisation without termination or transfer: all chains start together and keep growing as long as monomer remains, and they resume when more monomer is added. The anionic polymerisation of styrene initiated by butyllithium in a dry, aprotic solvent is the classic example.

Proposition 29.14 (Narrow distributions)

In a living polymerisation in which all chains start at once, the degree of polymerisation follows a Poisson distribution, and $\text{\DH} = 1 + \nu/(1+\nu)^2 \approx 1 + 1/\bar X_n$ , where $\nu$ is the mean number of monomers added per chain.

Proof. Admitted at this level. ∎

The distribution is that of the number of additions, independent random events at a common rate, made by each chain during the same time; the dispersity follows from the mean and variance of a Poisson law, both equal to $\nu$ . A living chain of 100 units thus has $\text{\DH} \approx 1.01$ , against nearly 2 for the same average made by step growth.

Two samples with the same X_n = 50 (models). Living chains crowd around 50 (1.02); step-growth chains spread from monomer to several hundred units (1.98). — Two samples with the same $\bar X_n = 50$ (models). Living chains crowd around 50 ( $\text{\DH} \approx 1.02$ ); step-growth chains spread from monomer to several hundred units ( $\text{\DH} \approx 1.98$ ).

Definition 29.15 (Copolymer)

A copolymer is a polymer built from two or more different monomers. Its units may follow each other at random (statistical copolymer), alternately, in long runs of each (block copolymer, made by living polymerisation), or as side chains of one grafted onto a backbone of the other (graft copolymer).

29.3 Structure of the chains

Definition 29.16 (Tacticity)

In a vinyl polymer $\ce{-[CH2-CHR]_n-}$ , every $\ce{CHR}$ carbon is a stereocentre. The tacticity describes their relative configurations along the chain drawn as a planar zig-zag: in an isotactic chain all the $\ce{R}$ groups lie on the same side of the plane, in a syndiotactic chain they alternate, and in an atactic chain they are placed at random.

Polypropene drawn as a zig-zag in the plane of the page; each methyl group is a wedge (towards the reader) or a hashed wedge (away). Isotactic: all methyls on one side; syndiotactic: alternating; atactic: irregular.

Regular chains can pack side by side into crystalline regions; irregular ones cannot. Isotactic polypropene, made with the metal catalysts of the Year 3 volume, is partly crystalline, hard and high melting; atactic polypropene is a soft, sticky amorphous material. A polymer is rarely fully crystalline: crystallites are embedded in amorphous regions where chains are tangled.

Definition 29.17 (Thermal transitions)

The degree of crystallinity of a polymer is the mass fraction of it that lies in crystalline regions. The glass transition temperature $T_g$ is the temperature below which the amorphous regions are a rigid glass, their chain segments unable to move, and above which they become rubbery; the crystalline regions melt at a higher temperature, $T_m$ .

Polystyrene and poly(methyl methacrylate) are amorphous with $T_g$ above room temperature: rigid, transparent glasses. Natural rubber has $T_g$ well below room temperature. Polyethene, with $T_g$ far below room temperature but highly crystalline, is tough and flexible: its crystallites hold the rubbery amorphous parts together.

Modulus (stiffness) of an amorphous polymer against temperature, schematic. Below T_g the material is a glass; through the transition the modulus falls by orders of magnitude; on the rubbery plateau entangled chains behave as a rubber; at higher temperature a linear polymer flows (solid line), while a cross-linked one keeps its plateau until it decomposes (dashed). — Modulus (stiffness) of an amorphous polymer against temperature, schematic. Below $T_g$ the material is a glass; through the transition the modulus falls by orders of magnitude; on the rubbery plateau entangled chains behave as a rubber; at higher temperature a linear polymer flows (solid line), while a cross-linked one keeps its plateau until it decomposes (dashed).

29.4 Properties and uses

Definition 29.18 (Elastomer)

An elastomer is a polymer used above its glass transition temperature whose chains are lightly cross-linked: it can be stretched to several times its length and returns to its shape when released.

The four large classes of polymer materials follow from structure and transitions. Thermoplastics, linear or branched, glassy or semi-crystalline at room temperature, soften on heating and can be moulded again and again: polyethene, polypropene, polystyrene, PET. Elastomers are lightly cross-linked chains above their $T_g$ : vulcanised rubber, in which sulfur bridges link the chains. Fibres are chains aligned by drawing, with strong interactions between them: the hydrogen bonds between amide groups in nylon. Thermosets are densely cross-linked networks formed during moulding: epoxy and phenol–formaldehyde resins, which cannot be remelted. Only thermoplastics are recycled by melting; the school volume gave the scale of the problem.

Polymer pellets in a plant, the form in which thermoplastics are sold, and fibres being drawn from a spinneret onto rotating rolls behind them (illustration).

History — A 99 % yield is not enough

Wallace Carothers, a young university instructor recruited into an industrial laboratory for fundamental research, built long molecules from well-known reactions, and his results strongly supported the then-contested idea that polymers are ordinary molecules, only very long. His team made polyesters, then, from 1934, polyamides; the equation that bears his name explains why only extreme conversions gave fibres. Nylon went into production in 1939. (Photograph: unknown photographer, public domain; Wikimedia Commons.)

Safety

Styrene is flammable and a health hazard. Dibenzoyl peroxide is an explosive oxidiser when dry and is kept damp and cool. Hexane-1,6-diamine and hexanedioic acid are corrosive to the eyes.

29.5 Exercises

Exercise 29.1 ★

Draw the repeat units of polyethene, polypropene, poly(vinyl chloride), polystyrene and nylon-6,6.

Solution

Solution of Exercise 29.1.

Polyethene $\ce{-[CH2-CH2]-}$ ; polypropene $\ce{-[CH2-CH(CH3)]-}$ ; poly(vinyl chloride) $\ce{-[CH2-CHCl]-}$ ; polystyrene $\ce{-[CH2-CH(C6H5)]-}$ ; nylon-6,6 $\ce{-[NH(CH2)6NH-CO(CH2)4CO]-}$ .

Exercise 29.2 ★

Step or chain growth: PET from ethane-1,2-diol and benzene-1,4-dicarboxylic acid; polystyrene; nylon-6,6; poly(methyl methacrylate); poly(lactic acid) from 2-hydroxypropanoic acid?

Solution

Solution of Exercise 29.2.

PET: step growth (polyester, water released). Polystyrene: chain growth. Nylon-6,6: step growth. Poly(methyl methacrylate): chain growth (a $\ce{C=C}$ monomer). Poly(lactic acid) from the hydroxy acid: step growth, an $\ce{A-B}$ monomer.

Exercise 29.3 ★

A polystyrene has $\bar X_n = 1000$ . Compute $\bar M_n$ , neglecting the end groups.

Solution

Solution of Exercise 29.3.

$\bar M_n = 1000 \times 104.15\,\mathrm{g}/\mathrm{mol} \approx 104\,\mathrm{kg}/\mathrm{mol}$ (styrene $\ce{C8H8}$ , $104.152\,\mathrm{g}/\mathrm{mol}$ ).

Exercise 29.4 ★

A polypropene chain drawn as a planar zig-zag has its methyl groups on the side of the reader, then away, then towards, then away, and so on. Name its tacticity. Could it crystallise?

Solution

Solution of Exercise 29.4.

Syndiotactic. Yes: a regular chain can pack into crystallites.

Exercise 29.5 ★★

A sample is a mixture of equal masses of two fractions, of molar masses $10\,\mathrm{kg}/\mathrm{mol}$ and $100\,\mathrm{kg}/\mathrm{mol}$ . Compute $\bar M_n$ , $\bar M_w$ and the dispersity.

Solution

Solution of Exercise 29.5.

Mass fractions 0.5 and 0.5. $1/\bar M_n = 0.5/10 + 0.5/100 = 0.055$ , $\bar M_n \approx 18.2\,\mathrm{kg}/\mathrm{mol}$ ; $\bar M_w = 0.5 \times 10 + 0.5 \times 100 = 55\,\mathrm{kg}/\mathrm{mol}$ ; $\text{\DH} \approx 3.0$ .

Exercise 29.6 ★★

Compute $\bar X_n$ for a balanced step-growth polymerisation at $p = 0.98$ and $p = 0.995$ .

Solution

Solution of Exercise 29.6.

$1/(1-0.98) = 50$ ; $1/(1-0.995) = 200$ .

Exercise 29.7 ★★

A diacid and a diamine are mixed with a 1 % excess (in moles) of the diamine. What is the largest $\bar X_n$ that can be reached?

Solution

Solution of Exercise 29.7.

$r = 1/1.01$ ; at $p = 1$ , $\bar X_n = (1+r)/(1-r) = (1.01 + 1)/(1.01 - 1) = 201$ .

Exercise 29.8 ★★

In a radical polymerisation the initiator concentration is doubled. By what factor do the rate of polymerisation and the kinetic chain length change?

Solution

Solution of Exercise 29.8.

$R_p \propto [\mathrm I]^{1/2}$ : multiplied by $\sqrt2 \approx 1.41$ . $\nu \propto [\mathrm I]^{-1/2}$ : divided by $\sqrt 2$ , about 0.71 times its former value.

Exercise 29.9 ★★

Poly(vinyl chloride) is rigid; garden hoses made of it are flexible because they contain a plasticiser, a small molecule mixed between the chains. Explain with the glass transition.

Solution

Solution of Exercise 29.9.

Pure PVC has its glass transition above room temperature: a rigid glass. The small plasticiser molecules sit between the chains, separate them and let segments move at lower temperature: $T_g$ falls below room temperature, and the material is rubbery and flexible in use.

Exercise 29.10 ★★★

Show that the Flory mass distribution $w_x = x(1-p)^2p^{\,x-1}$ , viewed as a function of a continuous $x$ , is largest at $x = -1/\ln p$ , and that this is close to $\bar X_n$ when $p$ is close to 1.

Solution

Solution of Exercise 29.10.

$\dfrac{\mathrm d}{\mathrm dx}\bigl(x p^{\,x-1}\bigr) = p^{\,x-1}(1 + x\ln p)$ , zero for $x = -1/\ln p$ (a maximum: the sign goes from $+$ to $-$ ). For $p$ close to 1, $\ln p \approx -(1-p)$ , so $x \approx 1/(1-p) = \bar X_n$ .

Exercise 29.11 ★★★

PET is made in two stages: dimethyl benzene-1,4-dicarboxylate with excess ethane-1,2-diol gives bis(2-hydroxyethyl) benzene-1,4-dicarboxylate and methanol; then this diester is heated under vacuum and polymerises, releasing ethane-1,2-diol. Write both stages and explain how the second stage solves the stoichiometry problem of the Carothers equation.

Solution

Solution of Exercise 29.11.

Stage 1, transesterification: $\ce{C6H4(COOCH3)2 + 2HOCH2CH2OH -> C6H4(COOCH2CH2OH)2 + 2CH3OH}$ , methanol distilled off. Stage 2: each diester molecule carries two $\ce{CH2CH2OH}$ ends; one end attacks an ester of another molecule and releases a molecule of ethane-1,2-diol, which is pumped off. The monomer carries both kinds of group in the right ratio, as an $\ce{A-B}$ monomer does: the balance $r = 1$ is built in, and removing the diol drives $p$ towards 1.

Exercise 29.12 ★★★

In a radical copolymerisation of monomers 1 and 2, the instantaneous composition of the copolymer is given by $\dfrac{\mathrm d[\mathrm M_1]}{\mathrm d[\mathrm M_2]} = \dfrac{[\mathrm M_1](r_1[\mathrm M_1] + [\mathrm M_2])}{[\mathrm M_2]([\mathrm M_1] + r_2[\mathrm M_2])}$ , where $r_1$ and $r_2$ are the reactivity ratios (exercise data: $r_1 = 2.0$ , $r_2 = 0.5$ ). Compute the ratio of units in the copolymer formed from an equimolar feed. What happens if $r_1 = r_2 = 0$ ?

Solution

Solution of Exercise 29.12.

Equimolar feed: ratio $= (2.0 + 1)/(1 + 0.5) = 2$ : twice as many units of monomer 1. With $r_1 = r_2 = 0$ , ratio $= [\mathrm M_1][\mathrm M_2]/([\mathrm M_2][\mathrm M_1]) = 1$ whatever the feed: each radical adds only the other monomer, an alternating copolymer.

29.6 Problem: Nylon-6,6 by the Kilogram

Problem 29.1

Weekend problem — the monomers and the nylon salt, the Carothers equation, a deliberate imbalance, the Flory distribution and the water released

Nylon-6,6 is made from hexane-1,6-diamine and hexanedioic acid. End groups are neglected throughout. Molar masses ( $\mathrm{g}/\mathrm{mol}$ ): C 12.011, H 1.008, N 14.007, O 15.999.

Part I — Monomers and repeat unit.

Write the formulas of the two monomers.
Write the equation of the formation of one amide link.
The monomers are first combined as a 1 : 1 salt, crystallised from solution. Why?
Give the repeat unit of nylon-6,6, its formula and its molar mass.
What is the mean molar mass $M_0$ of a monomer-derived unit in the chain?
What do the two numbers in the name “6,6” count?

Part II — The Carothers equation.

Compute $\bar X_n$ and $\bar M_n$ at $p = 0.98$ .
The same at $p = 0.99$ .
Explain the remark “a 99 % yield is not enough”.
Which extent of reaction gives $\bar M_n = 10\,\mathrm{kg}/\mathrm{mol}$ ?
How is the extent pushed so high in practice?
Why must the monomers be very pure?

Part III — A deliberate imbalance.

The diamine is used in 1 % molar excess. Compute $r$ .
Compute the largest $\bar X_n$ and $\bar M_n$ that can then be reached.
Compute $\bar X_n$ at $p = 0.99$ with this imbalance.
Which groups end the chains at the end of the reaction?
A little ethanoic acid is sometimes added. What does it do to the chains?
Why would a manufacturer limit the chain length on purpose?

Part IV — Distribution and mass balance.

For a balanced mixture at $p = 0.99$ , give the dispersity.
Compute $\bar M_w$ at $p = 0.99$ .
At $p = 0.99$ , what fraction of the molecules, and what fraction of the mass, is still unreacted monomer?
Near which chain length is the mass distribution largest?
Compute the mass of water released per kilogram of nylon-6,6.
Why does melt spinning need $\bar M_n$ within a window?
State the extent of reaction that gives $\bar M_n = 20\,\mathrm{kg}/\mathrm{mol}$ for a balanced mixture.

Solution

Solution of Problem 29.1.

1. $\ce{H2N(CH2)6NH2}$ and $\ce{HOOC(CH2)4COOH}$ . 2. One water per link:

\mathrm{{-}COOH + H_2N{-} \longrightarrow {-}CO{-}NH{-} + H_2O}.

3. The salt contains exactly one diamine per diacid; weighing two separate liquids or solids would never reach the exactness the Carothers equation demands. 4. $\ce{-[NH(CH2)6NH-CO(CH2)4CO]-}$ , $\ce{C12H22N2O2}$ , $226.32\,\mathrm{g}/\mathrm{mol}$ . 5. The repeat unit contains two monomer-derived units: $M_0 = 226.32/2 = 113.16\,\mathrm{g}/\mathrm{mol}$ . 6. The carbons of the diamine (6) and of the diacid (6). 7. $\bar X_n = 50$ , $\bar M_n = 50 \times 113.16 \approx 5.66\,\mathrm{kg}/\mathrm{mol}$ . 8. $\bar X_n = 100$ , $\bar M_n \approx 11.3\,\mathrm{kg}/\mathrm{mol}$ . 9. With 99 % of the groups reacted, the average chain has only 100 units, too short for a strong fibre; each further step towards complete conversion doubles or triples the chain length. 10. $\bar X_n = 10\,000/113.16 = 88.4$ ; $p = 1 - 1/88.4 \approx 0.989$ . 11. By removing water: the melt is heated well above its melting point, finally under reduced pressure, so that the equilibrium keeps moving towards amide. 12. A monofunctional impurity caps chain ends, and any impurity upsets the 1 : 1 balance. 13. $r = 1/1.01 \approx 0.990$ . 14. $\bar X_n = (1+r)/(1-r) = 201$ ; $\bar M_n \approx 201 \times 113.16 \approx 22.7\,\mathrm{kg}/\mathrm{mol}$ . 15. $\bar X_n = 1.9901/(1.9901 - 2 \times 0.9901 \times 0.99) \approx 1.9901/0.0297 \approx 67$ . 16. Amine groups: when the acid groups are used up, every chain ends in $\ce{NH2}$ . 17. It turns an amine end into an amide that cannot react further: it caps chains, limiting $\bar M_n$ and stopping further growth when the polymer is remelted. 18. To keep a reproducible melt viscosity for spinning and moulding, which would otherwise drift as chains keep reacting in the melt. 19. $\text{\DH} = 1 + p = 1.99$ . 20. $\bar M_w = 1.99 \times 11.3 \approx 22.5\,\mathrm{kg}/\mathrm{mol}$ . 21. Number fraction $n_1 = 1 - p = 0.01$ , 1 % of the molecules; mass fraction $w_1 = (1-p)^2 = 10^{-4}$ , 0.01 % of the mass. 22. Near $x = -1/\ln 0.99 \approx 99.5$ , that is near $\bar X_n = 100$ (exercise 10). 23. $1000/226.32 = 4.42\,\mathrm{mol}$ of repeat units, two amide links each: $8.84\,\mathrm{mol}$ of water, $8.84 \times 18.015 \approx 159\,\mathrm{g}$ . 24. Chains too short give a weak fibre; chains too long give a melt too viscous to be pushed through the fine holes of the spinneret. 25. $\bar X_n = 20\,000/113.16 = 176.7$ and $p = 1 - 1/176.7 = \boldsymbol{\approx 0.994}$ .

Terms defined in this chapter

See all 852 terms in the glossary