Condensation polymerisation
A polyester T-shirt, a nylon rope and a fleece jacket are all made from monomers that join end to end and give off a small molecule every time they link. That is very different from addition polymerisation, where a double bond simply opens. Here you have to keep track of what is lost as well as what is joined.
Condensation polymerisation: joining with loss of a small molecule
Imagine heating terephthalic acid with ethane-1,2-diol in an open flask in the school lab. The total mass falls, a colourless liquid distils off and the residue thickens into PET fibres. The distillate is water. Where did it come from, when the reactants were pure and dry?
You met esterification when studying alcohols and carboxylic acids. Condensation polymerisation is the same reaction repeated many times.
In the synthesis of a linear polyester, ethanedioic acid ($\mathrm{HOOC-COOH}$) is reacted with ethane-1.2-diol ($\mathrm{HO-CH_2-CH_2-OH}$). a) Write the balanced…
In the synthesis of a linear polyester, ethanedioic acid ($\mathrm{HOOC-COOH}$) is reacted with ethane-1.2-diol ($\mathrm{HO-CH_2-CH_2-OH}$).
- a) Write the balanced chemical equation for the formation of the simplest linear dimer formed from one molecule of each reactant, identifying the new functional group formed and the by-product eliminated.
- b) Justify why both monomers must be bifunctional to allow indefinite growth of the polymer chain.
Proteins are natural biopolymers whose formal formation balance corresponds to the condensation of $\alpha$-amino acids of general formula $\mathrm{H_2N-CHR-COOH}$. a)…
Proteins are natural biopolymers whose formal formation balance corresponds to the condensation of $\alpha$-amino acids of general formula $\mathrm{H_2N-CHR-COOH}$.
- a) Explain the nature of the bond joining the monomers and write the general condensation reaction between two molecules of glycine ($\mathrm{H_2N-CH_2-COOH}$), indicating the molecule that is released.
- b) Explain why it is not possible to chemically synthesise in the laboratory a complex cellular protein with a specific sequence simply by heating a mixture of its monomers in solution.
Repeating unit and molar mass
For nylon 6,6, the repeating unit is $-\mathrm{OC-(CH_2)_4-CO-NH-(CH_2)_6-NH}-$. For PET it is $-\mathrm{OC-C_6H_4-CO-O-CH_2CH_2-O}-$. A hydroxy acid such as lactic acid has both groups in one molecule, so each monomer contributes one unit and loses only one water: $M_{\text{unit}} = M_{\text{monomer}} - M(\mathrm{H_2O})$.
Poly(ethylene terephthalate), known commercially as PET, is a polyester synthesised by condensation from benzene-1.4-dicarboxylic acid (terephthalic acid,…
Poly(ethylene terephthalate), known commercially as PET, is a polyester synthesised by condensation from benzene-1.4-dicarboxylic acid (terephthalic acid, $\mathrm{C_8H_6O_4}$, $M = 166.13\,\text{g}\times\text{mol}^{-1}$) and ethane-1.2-diol ($\mathrm{C_2H_6O_2}$, $M = 62.07\,\text{g}\times\text{mol}^{-1}$).
- a) Deduce the formula of the repeating unit corresponding to one diacid/diol pair of the polymer in the very-long-chain limit.
- b) Calculate the molar mass of this repeating unit, given that in each cycle of incorporation of a diacid/diol pair two water molecules are eliminated ($\mathrm{H_2O}$, $M = 18.02\,\text{g}\times\text{mol}^{-1}$).
Nylon 6.6 is a polyamide synthesised industrially by the reaction of hexanedioic acid ($\mathrm{HOOC-(CH_2)4-COOH}$, $M = 146.14\,\text{g}\times\text{mol}^{-1}$) and…
Nylon 6.6 is a polyamide synthesised industrially by the reaction of hexanedioic acid ($\mathrm{HOOC-(CH_2)_4-COOH}$, $M = 146.14\,\text{g}\times\text{mol}^{-1}$) and hexane-1.6-diamine ($\mathrm{H_2N-(CH_2)_6-NH_2}$, $M = 116.21\,\text{g}\times\text{mol}^{-1}$).
- a) Write the structure of the repeating unit of the polymer in the long-chain limit and calculate its theoretical molar mass, given that $M(\mathrm{H_2O}) = 18.02\,\text{g}\times\text{mol}^{-1}$.
- b) A reactor produces a linear oligomer from exactly 10 diacid molecules and 10 diamine molecules with uncyclised terminal ends ($-\mathrm{COOH}$ and $-\mathrm{NH_2}$). Calculate the exact molar mass of this finite oligomer.
Poly(lactic acid) (PLA) is a biodegradable polyester that can be obtained by self-condensation of 2-hydroxypropanoic acid (lactic acid, $\mathrm{CH_3-CH(OH)-COOH}$,…
Poly(lactic acid) (PLA) is a biodegradable polyester that can be obtained by self-condensation of 2-hydroxypropanoic acid (lactic acid, $\mathrm{CH_3-CH(OH)-COOH}$, $M = 90.08\,\text{g}\times\text{mol}^{-1}$), a bifunctional hydroxycarboxylic monomer.
- a) Represent the semi-structural formula of the repeating unit of PLA in a linear polymer chain and calculate its molar mass, taking $M(\mathrm{H_2O}) = 18.02\,\text{g}\times\text{mol}^{-1}$.
- b) A homogeneous sample is polymerised to give linear chains formed by 150 monomer units of lactic acid. Calculate the exact molar mass of these chains, taking the terminal ends as intact ($-\mathrm{H}$ and $-\mathrm{OH}$).
Finite chains: counting monomers and water molecules
In a lab synthesis of a polyamide, joining 2 monomers gives 1 water molecule, joining 3 gives 2 and joining 4 gives 3. Why is the amount of water always one less than the number of monomers joined?
For the molar mass of a finite chain, add the masses of all the monomers and subtract $(N-1) \times 18.02$. The end groups are kept, so this is not the same as $N$ repeating units.
Step-growth condensation between bifunctional monomers generates oligomers before high molecular masses are reached. The following diagram illustrates the formation of a…
Step-growth condensation between bifunctional monomers generates oligomers before high molecular masses are reached. The following diagram illustrates the formation of a linear oligomer from two molecules of a dicarboxylic acid $\mathrm{HOOC-R_1-COOH}$ and two molecules of a diol $\mathrm{HO-R_2-OH}$ in alternation:
- a) From the linear diagram of 4 connected monomer blocks, determine the exact number of water molecules released during the formation of this tetramer and write the functional groups left free at each end of the chain.
- b) If an open linear chain is formed by the strict polycondensation of $n$ diacid molecules and $n$ diol molecules without undergoing cyclisation, deduce the general expression for the number of water molecules generated in terms of $n$.
In the synthesis of nylon 6,6, $146.14\,\text{g}$ of hexanedioic acid ($1.00\,\text{mol}$, $M = 146.14\,\text{g}\times\text{mol}^{-1}$) and $116.21\,\text{g}$ of…
In the synthesis of nylon 6,6, $146.14\,\text{g}$ of hexanedioic acid ($1.00\,\text{mol}$, $M = 146.14\,\text{g}\times\text{mol}^{-1}$) and $116.21\,\text{g}$ of hexane-1.6-diamine ($1.00\,\text{mol}$, $M = 116.21\,\text{g}\times\text{mol}^{-1}$) are placed in a reactor. The reaction is stopped, without cyclisation, on reaching a mean chain length of $n = 50$ diacid-diamine pairs per macromolecule (degree of polymerisation in number of monomers $\bar{X}_n = 100$). Data: $M(\mathrm{H_2O}) = 18.02\,\text{g}\times\text{mol}^{-1}$.
- a) Determine the total moles and the exact mass of water generated during condensation in the whole reactor.
- b) Calculate the number-average molecular mass ($\bar{M}_n$) of the resulting polymer, taking the end groups into account, and determine the percentage error made if it were calculated approximately as $50\times M_{\text{rep}}$.
Mass balance in industry, and biopolymers
An industrial batch of PET is manufactured from terephthalic acid ($\mathrm{C_8H_6O_4}$, $M = 166.13\,\text{g}\times\text{mol}^{-1}$) and ethane-1.2-diol…
An industrial batch of PET is manufactured from terephthalic acid ($\mathrm{C_8H_6O_4}$, $M = 166.13\,\text{g}\times\text{mol}^{-1}$) and ethane-1.2-diol ($\mathrm{C_2H_6O_2}$, $M = 62.07\,\text{g}\times\text{mol}^{-1}$) in strict equimolar ratio ($10.00\,\text{mol}$ of each reactant).
- a) If the process ends when the number-average degree of polymerisation of the linear mixture reaches $\bar{X}_n = 200$ (that is, chains of 200 monomers in total, 100 of each type, uncyclised), calculate the mass of dry polyester obtained and the mass yield of the polymer relative to the initial mass of reactants.
- b) Justify which thermodynamic factor makes it necessary in industry to continuously remove the water vapour generated, under vacuum and at high temperature. Data: $M(\mathrm{H_2O})=18.02\,\mathrm{g}\times\mathrm{mol}^{-1}$.
A linear tetrapeptide with the sequence $\mathrm{H_2N-Ala-Gly-Ala-Gly-COOH}$ is synthesised from $L$-alanine ($\mathrm{C_3H_7NO_2}$,…
A linear tetrapeptide with the sequence $\mathrm{H_2N-Ala-Gly-Ala-Gly-COOH}$ is synthesised from $L$-alanine ($\mathrm{C_3H_7NO_2}$, $M = 89.09\,\text{g}\times\text{mol}^{-1}$) and glycine ($\mathrm{C_2H_5NO_2}$, $M = 75.07\,\text{g}\times\text{mol}^{-1}$).
- a) Write the stoichiometric reaction for the formation of the tetrapeptide, calculate its exact molar mass given that $M(\mathrm{H_2O}) = 18.02\,\text{g}\times\text{mol}^{-1}$, and calculate the volume that the water from the formal condensation balance of $0.250\,\text{mol}$ of tetrapeptide would occupy as vapour, measured separately at $120.0\,^\circ\text{C}$ and $1.00\,\text{atm}$ ($R = 0.082\,\text{atm}\times\text{L}\times\text{mol}^{-1}\times\text{K}^{-1}$).
- b) Explain why natural proteins have unique catalytic and structural properties that cannot be reproduced by the simple random condensation polymerisation of their constituent monomers.
Back to the start: to predict the mass of the finished T-shirt fibre or nylon rope, you cannot just add up the masses of the reactants. You must subtract $18.02\ \text{g mol}^{-1}$ for every link formed, because that mass leaves as water vapour.