IChO 2002 · Theory problem 5

Aramids, High-performance Polymeric Materials

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Aromatic polyamides (aramids) are high strength, high performance polymer fibers that find use in composite materials, bullet-proof vests, high quality skis, safety helmets, etc. Aramid PPTA is marketed under the names Kevlar® (DuPont) and Twaron® (Teijin), and amongst others manufactured in the north of The Netherlands. The PPTA chains are neatly packed into fibers with a sheet type structure.

Figure 1

PPTA

5.1 Draw the structure of these sheets (three chains suffice).

For a polymerisation of equimolar amounts of two monomers the average chain length is \(\overline{P}_n\), the degree of conversion is \(p\), which equals the fraction of functional groups that have reacted, the total number of chains is \(N_t\) and the total initial number of monomers is \(U_0\).

Assuming that the polymerization equilibrium can fully be described by:

Figure 2

where C stands for any \(\ce{-CO2}\) group, A stands for any \(\ce{-NH2}\) group and Am stands for any amide group.

5.2 Calculate the degree of conversion needed to obtain an average chain length of 500.

5.3 For the synthesis of PPTA the following possibilities are considered. Which of the following reactions will work? Mark the correct answer(s).

Figure 3

5.4 Another type of aramid can be produced from 4-aminobenzoic acid (4-aminobenzene-carboxylic acid) by heating.

(a) Give the structure of this aramid (\(n=4\))

(b) Calculate the average chain length at equilibrium (reaction is carried out in a closed vessel). The equilibrium constant \(K=576\).

Official English text of the problem.

Answers

  • 5.1: Three extended, parallel PPTA chains joined into a sheet by repeated interchain \(\ce{N-H}\cdots\ce{O=C}\) hydrogen bonds. The repeat-fragment SMILES and drawing description are given below.
  • 5.2: \(\boxed{p=0.998=99.8\%}\).
  • 5.3: Mark the second and third reactions: terephthaloyl chloride with the diamine, and heated dimethyl terephthalate with the diamine.
  • 5.4(a): The linear poly(p-benzamide) tetramer: \[ \ce{H2N-R-CO-NH-R-CO-NH-R-CO-NH-R-COOH}, \] where \(R\) is para-phenylene. Its explicit SMILES is given below.
  • 5.4(b): \(\boxed{\overline{P}_n=25}\) monomer residues per chain, with equilibrium conversion \(p=0.960\).

Solution

5.1 — Hydrogen-bonded sheets

PPTA is poly(p-phenylene terephthalamide). Its repeating fragment has SMILES
[*]Nc1ccc(NC(=O)c2ccc(C(=O)[*])cc2)cc1.

The two [*] atoms indicate continuation into neighbouring repeat units; this notation describes the covalent connectivity.

To draw the requested sheet:

  1. Draw three extended PPTA chains with parallel chain axes and predominantly trans amide groups.
  2. Align adjacent chains so that an amide \(\ce{N-H}\) group directed into the gap faces a carbonyl oxygen on the neighbouring chain.
  3. Draw a dashed hydrogen bond from that hydrogen to the oxygen: \[ \ce{N-H}\cdots\ce{O=C}. \]
  4. Repeat these contacts along both gaps, so that the middle chain hydrogen-bonds to both outer chains.

The nearly linear donor–hydrogen–acceptor contacts hold the chains together within the sheet. Their spatial arrangement and the amide conformations are not specified by the SMILES.

5.2 — Conversion required for a chain length of 500

Each monomer is bifunctional, so the initial number of functional groups is \(2U_0\). At conversion \(p\), the number of reacted functional groups is \(2U_0p\).

Each new amide linkage consumes two functional groups. Consequently, the number of linkages formed is \(U_0p\). For linear polymerization without cyclization, each linkage reduces the number of separate molecules by one:

\[ N_t=U_0-U_0p=U_0(1-p). \]

The number-average chain length therefore obeys the Carothers equation:

\[ \overline{P}_n=\frac{U_0}{N_t}=\frac{1}{1-p}. \]

Thus,

\[ 500=\frac{1}{1-p} \quad\Longrightarrow\quad p=1-\frac{1}{500} =\boxed{0.998=99.8\%}. \]

Here chain length counts individual original monomer residues.

5.3 — Suitable synthesis routes

The common diamine is p-phenylenediamine, also called benzene-1,4-diamine, SMILES Nc1ccc(N)cc1.

Scheme, from top to bottom Carbonyl-containing monomer and SMILES Works under the indicated conditions? Reason
1 Terephthalic acid, O=C(O)c1ccc(C(=O)O)cc1 No At \(25\,^\circ\mathrm C\), the unactivated carboxylic acid and amine primarily undergo acid–base interaction. These conditions do not provide effective dehydrative amidation.
2 Terephthaloyl chloride, O=C(Cl)c1ccc(C(=O)Cl)cc1 Yes Acid chlorides readily acylate amines. The base captures the acid produced and maintains the amine’s nucleophilicity.
3 Dimethyl terephthalate, COC(=O)c1ccc(C(=O)OC)cc1 Yes Heating enables ester aminolysis: amide bonds form with elimination of methanol.
4 Di-tert-butyl terephthalate, CC(C)(C)OC(=O)c1ccc(C(=O)OC(C)(C)C)cc1 No The hindered, unactivated tert-butyl ester does not undergo effective aminolysis with this aromatic diamine at \(30\,^\circ\mathrm C\).

For balanced polymerization equations, let \(R\) denote para-phenylene, or benzene-1,4-diyl, fragment SMILES [*]c1ccc([*])cc1. In the condensed structures below, \(\ce{CO}\) denotes a carbonyl group, \(\ce{C(=O)}\).

A representative finite chain containing \(n\) molecules of each monomer has \(2n-1\) newly formed amide bonds. The acid-chloride route is therefore

\[ \begin{aligned} &\ce{n ClCO-R-COCl + n H2N-R-NH2}\\ &\qquad\ce{-> H-[NH-R-NH-CO-R-CO]_{n}-Cl + (2n-1) HCl}. \end{aligned} \]

The corresponding methyl-ester route is

\[ \begin{aligned} &\ce{n CH3OCO-R-COOCH3 + n H2N-R-NH2}\\ &\qquad\ce{->[\Delta] H-[NH-R-NH-CO-R-CO]_{n}-OCH3 + (2n-1) CH3OH}. \end{aligned} \]

These equations retain the end groups explicitly: the leading hydrogen completes an amino end group, while the other end retains an acid chloride or methyl ester group.

The byproducts are hydrogen chloride, SMILES Cl, and methanol, SMILES CO, respectively. In the second scheme, the added base captures the hydrogen chloride.

5.4(a) — Aramid from 4-aminobenzoic acid

The monomer is 4-aminobenzoic acid, SMILES Nc1ccc(C(=O)O)cc1. Its amino group reacts with another molecule’s carboxyl group, producing poly(p-benzamide).

Four monomers form three amide bonds and release three molecules of water, SMILES O:

\[ \begin{aligned} &\ce{4 H2N-R-COOH}\\ &\qquad\ce{->[\Delta] H2N-R-CO-NH-R-CO-NH-R-CO-NH-R-COOH + 3 H2O}. \end{aligned} \]

The linear 4-aminobenzoic acid tetramer has SMILES
Nc1ccc(C(=O)Nc2ccc(C(=O)Nc3ccc(C(=O)Nc4ccc(C(=O)O)cc4)cc3)cc2)cc1.

The requested drawing should show four benzene rings, each substituted in the para positions, connected successively by three \(\ce{-C(=O)-NH-}\) links. One end is an amino group and the other is a carboxylic acid group. Its molecular formula is \(\ce{C28H22N4O5}\).

5.4(b) — Equilibrium chain length in a closed vessel

Use the stated functional-group equilibrium:

\[ \ce{C + A <=> Am + H2O}. \]

Let \(c_0\) be the initial monomer concentration, in \(\mathrm{mol\,L^{-1}}\). Taking the starting monomer to contain no water, each amide linkage produces one water molecule, which remains in the closed vessel.

At conversion \(p\),

\[ [\mathrm C]=[\mathrm A]=c_0(1-p), \qquad [\mathrm{Am}]=[\ce{H2O}]=c_0p. \]

Under the problem’s equilibrium model,

\[ K= \frac{[\mathrm{Am}][\ce{H2O}]}{[\mathrm C][\mathrm A]} = \frac{(c_0p)^2}{[c_0(1-p)]^2} = \left(\frac{p}{1-p}\right)^2. \]

Therefore,

\[ \frac{p}{1-p}=\sqrt{576}=24, \qquad p=\frac{24}{25}=0.960. \]

Applying the chain-count relation from 5.2,

\[ \boxed{\overline{P}_n=\frac{1}{1-p}=25}. \]

Thus the equilibrium average is 25 monomer residues per chain.

The solution was checked against the official solution.