IChO 2010 · Theory problem 8

Polycondensation and the degree of polymerization

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The esterification reaction between bi-functional molecules gives one of the typical linear chain polymers, as shown in equation (1), by polycondensation (often called “condensation polymerization”). The control of polymerization conditions and procedures determines the length of polymer strands, i. e. the average degree of polymerization, \(X\) (note that \(X=2n\) in the present instance). Because \(X\) (and also \(n\)) is an averaged number, it is not always an integer but a value with decimal figures.

Figure 1

\(X\) can be estimated from the consumption of functional groups (here, \(\ce{-COOH}\) and \(\ce{-OH}\)). Let us define the degree of reaction, \(p\), as \(p=(N_0-N)/N_0\;(\leq 1)\), where \(N_0\) and \(N\) denote the total numbers of functional groups before and after the polymerization, respectively. For each functional group of the dicarboxylic acid molecules (A) and diol molecules (B), we add the suffixes of “A” or “B” such as \(N_{\mathrm{A}0}\), \(N_{\mathrm{B}0}\), \(N_{\mathrm A}\) or \(N_{\mathrm B}\), respectively, i. e. \(N_0=N_{\mathrm{A}0}+N_{\mathrm{B}0}\) and \(N=N_{\mathrm A}+N_{\mathrm B}\). When the initial feed is unbalanced such as \(N_{\mathrm{A}0}\leq N_{\mathrm{B}0}\), \(X\) is expressed by \(p_{\mathrm A}\) and \(r\) as shown in equation 2, where \(r=N_{\mathrm{A}0}/N_{\mathrm{B}0}\;(\leq 1)\) and \(p_{\mathrm A}=(N_{\mathrm{A}0}-N_{\mathrm A})/N_{\mathrm{A}0}\). If \(r=1\), \(p_{\mathrm A}\) is identical to \(p\) and equation 2 becomes the same to the Carothers equation.

\[ X=\frac{1+r}{1+r-2p_{\mathrm A}r} \tag{2} \]

Some nylon-6,6 sample was prepared by polycondensation between an equimolar mixture of adipic acid (hexanedioic acid) and hexamethylenediamine (hexane-1,6-diamine).

8.1 Show the chemical structure of this nylon-6,6 sample.

[Caution: What are the end groups when polycondensation was started from the equimolar mixture?]

8.2 When an average molar weight, \(M\), of this nylon-6,6 sample is \(5507.25\;(\mathrm{g\,mol^{-1}})\), calculate its \(X\) value to the second decimal place.

8.3 Give the \(p\) value to the fifth decimal place that is necessary to prepare the above nylon-6,6 sample. If you get no numerical answer in 8.2 use the value 52.50 instead.

The low-molecular-weight polyester (oligoester) is prepared from the mixture of 36.54 g of adipic acid (hexanedioic acid) and an unknown amount [\(W\) (g)] of butane-1,4-diol (Bdiol). Under the condition of \(p_{\mathrm A}\to 1\), the oligoester with \(X=11.00\) carrying Bdiol units in both chain ends, is obtained.

8.4 Show the precise chemical structure of this oligoester of \(X=11.00\).

8.5 Calculate the unknown amount, \(W\) (g), to the first decimal place.

Official English text of the problem.

Answers

  • 8.1: Nylon-6,6, represented by \[ \boxed{\ce{HO-[CO(CH2)4CO-NH(CH2)6NH]_{n}-H}}, \] with carboxyl and amino end groups. Repeat-fragment SMILES: [*]C(=O)CCCCC(=O)NCCCCCCN[*]; the full-chain construction is specified below.
  • 8.2: \(\boxed{X=48.50}\), using the atomic-weight convention stated below.
  • 8.3: \(\boxed{p=0.97938}\).
  • 8.4: Hydroxyl-terminated oligo(butylene adipate), containing five adipate residues and six butane-1,4-diol residues. Exact 11-unit-chain SMILES: OCCCCOC(=O)CCCCC(=O)OCCCCOC(=O)CCCCC(=O)OCCCCOC(=O)CCCCC(=O)OCCCCOC(=O)CCCCC(=O)OCCCCOC(=O)CCCCC(=O)OCCCCO.
  • 8.5: \(\boxed{W=27.0\ \mathrm g}\).

Solution

For the numerical calculations, use the two-decimal atomic weights \[ A_r(\ce C)=12.01,\qquad A_r(\ce H)=1.01,\qquad A_r(\ce N)=14.01,\qquad A_r(\ce O)=16.00. \]

8.1

The monomers are:

  • Adipic acid (hexanedioic acid), \(\ce{HOOC(CH2)4COOH}\), SMILES O=C(O)CCCCC(=O)O.
  • Hexamethylenediamine (hexane-1,6-diamine), \(\ce{H2N(CH2)6NH2}\), SMILES NCCCCCCN.

A carboxyl group and an amino group condense to form an amide linkage, releasing water, \(\ce{H2O}\), SMILES O. Connecting \(n\) molecules of each monomer into one linear chain requires \(2n-1\) amide bonds. The balanced equation is

\[ n\,\ce{HOOC(CH2)4COOH} +n\,\ce{H2N(CH2)6NH2} \longrightarrow \ce{HO-[CO(CH2)4CO-NH(CH2)6NH]_{n}-H} +(2n-1)\,\ce{H2O}. \]

Thus, the requested representative nylon-6,6 chain has one terminal \(\ce{-COOH}\) group and one terminal \(\ce{-NH2}\) group. Its repeating connectivity is

\[ \ce{-C(=O)-(CH2)4-C(=O)-NH-(CH2)6-NH-}. \]

In the repeat-fragment SMILES given above, [*] marks attachment to adjacent units. For any positive integer \(n\), a full-chain SMILES with the displayed end groups is constructed by writing O followed by \(n\) consecutive copies of C(=O)CCCCC(=O)NCCCCCCN. A single ordinary molecular SMILES cannot encode a distribution of chain lengths or a noninteger average \(n\).

Strictly, an equimolar feed ensures equal total numbers of residual carboxyl and amino groups throughout the sample. It does not require every individual chain to have unlike end groups; the displayed chain represents the balanced composition.

8.2

The monomer and water molar masses are

\[ \begin{aligned} M_{\text{adipic acid}} &=6(12.01)+10(1.01)+4(16.00) =146.16\ \mathrm{g\,mol^{-1}},\\ M_{\text{diamine}} &=6(12.01)+16(1.01)+2(14.01) =116.24\ \mathrm{g\,mol^{-1}},\\ M_{\text{water}} &=2(1.01)+16.00 =18.02\ \mathrm{g\,mol^{-1}}. \end{aligned} \]

Using the balanced condensation equation,

\[ \begin{aligned} M &=n(146.16)+n(116.24)-(2n-1)(18.02)\\ &=226.36n+18.02 \qquad(\mathrm{g\,mol^{-1}}). \end{aligned} \]

The \(18.02\ \mathrm{g\,mol^{-1}}\) term accounts for the chain ends. Hence,

\[ n=\frac{5507.25-18.02}{226.36}=24.25. \]

Because \(X\) counts the residues of both monomers,

\[ \boxed{X=2n=48.50}. \]

The noninteger \(n\) is an average, as specified in the problem.

8.3

For the equimolar mixture, \(r=1\) and \(p_{\mathrm A}=p\). Equation (2) gives

\[ X=\frac{2}{2-2p}=\frac{1}{1-p}. \]

Therefore,

\[ p=1-\frac1X =1-\frac1{48.50} =0.979381443\ldots, \]

so, to five decimal places,

\[ \boxed{p=0.97938}. \]

8.4

The diol is butane-1,4-diol, \(\ce{HO(CH2)4OH}\), SMILES OCCCCO.

In the limit \(p_{\mathrm A}\to1\), all carboxyl groups are esterified. A linear chain ending in diol residues must alternate as

\[ B-A-B-A-\cdots-A-B. \]

If it contains \(a\) acid residues and \(b\) diol residues, then

\[ b=a+1,\qquad a+b=11. \]

Consequently,

\[ a=5,\qquad b=6. \]

The precise requested 11-unit structure is therefore

\[ \boxed{ \ce{HO-(CH2)4-[O-C(=O)-(CH2)4-C(=O)-O-(CH2)4]5-OH} }. \]

This hydroxyl-terminated oligo(butylene adipate) has five adipate residues alternating with six diol residues. Both ends are \(\ce{-CH2OH}\), and all ten carboxyl groups have become ester groups. Its full SMILES is given in Answers 8.4.

The balanced condensation equation is

\[ \begin{aligned} &5\,\ce{HOOC(CH2)4COOH} +6\,\ce{HO(CH2)4OH}\\ &\quad\longrightarrow \ce{HO-(CH2)4-[OCO(CH2)4COO(CH2)4]5-OH} +10\,\ce{H2O}. \end{aligned} \]

The product has molecular formula \(\ce{C54H90O22}\). Since \(X\) is an average, the sample may contain different chain lengths; the structure above is the requested 11-unit representative.

8.5

Both monomers are bifunctional, so their initial functional-group ratio equals their molar ratio:

\[ r=\frac{N_{\mathrm{A}0}}{N_{\mathrm{B}0}} =\frac{n_{\text{acid}}}{n_{\text{diol}}}. \]

At \(p_{\mathrm A}\to1\),

\[ 11.00=\frac{1+r}{1-r}, \]

which gives

\[ r=\frac{11-1}{11+1}=\frac56. \]

The amount of adipic acid is

\[ n_{\text{acid}} =\frac{36.54\ \mathrm g}{146.16\ \mathrm{g\,mol^{-1}}} =0.2500\ \mathrm{mol}. \]

Therefore,

\[ n_{\text{diol}}=\frac65(0.2500)=0.3000\ \mathrm{mol}. \]

For butane-1,4-diol,

\[ M_{\text{diol}} =4(12.01)+10(1.01)+2(16.00) =90.14\ \mathrm{g\,mol^{-1}}. \]

Thus,

\[ W=(0.3000\ \mathrm{mol})(90.14\ \mathrm{g\,mol^{-1}}) =27.042\ \mathrm g, \]

and to one decimal place,

\[ \boxed{W=27.0\ \mathrm g}. \]

The solution was checked against the official solution.