Prof. Otto Wichterle was a famous Czech polymer chemist and inventor of soft contact lenses. He also — Analytical Chemistry Chemistry Question
Ring opening polymerization (ROP)
Prof. Otto Wichterle was a famous Czech polymer chemist and inventor of soft contact lenses. He also contributed to the production of an industrially important polymer poly(ε-caprolactam) (silon, A) by ring opening polymerization of ε-caprolactam (hexano-6-lactam).
The polymerization reaction is usually carried out by a special type of anionic polymerization initiated by the addition of a small amount of acetic anhydride to an excess of ε-caprolactam. Compound B is formed which contains an imide bond that is more susceptible to nucleophilic attack than that of the amide bond in ε-caprolactam. The molar amount of B is the same as the molar amount of the subsequently formed polymer chains.
Draw the structure of compound B.
Model Answer
[VISUAL] N-acetylcaprolactam (N-acetyl-hexano-6-lactam) structure.
Write an arrow-pushing mechanism of the described initiation and propagation steps.
Model Answer
[VISUAL] Mechanism of initiation and propagation:
(1) Deprotonation of ε-caprolactam by base to form ε-caprolactam anion.
(2) Nucleophilic attack of ε-caprolactam anion on the imide carbonyl of compound B.
(3) Ring-opening of compound B.
(4) Protonation of the ring-opened intermediate by another ε-caprolactam molecule, resulting in an unreactive N-alkyl acetamide end and regenerating the ε-caprolactam anion.
(5) Propagation by nucleophilic attack of the regenerated ε-caprolactam anion on the activated imide end of the growing chain.
Draw the structure of poly(ε-caprolactone) prepared with sodium ethoxide as the initiator and water as the terminator.
Model Answer
[VISUAL] CH3CH2O-[C(=O)(CH2)5O]n-H
Two kilograms of ε-caprolactone were polymerized with 10 g sodium ethoxide with 83% conversion. Calculate the number-average molecular weight of the obtained polymer (use atomic masses of elements rounded to whole numbers). Neglect the weight contribution of the initiator residue to the molecular weight of the polymer.
Model Answer
Ten grams of sodium ethoxide correspond to 10 / (2 × 12 + 5 × 1 + 1 × 16 + 1 × 23) = 0.1471 mol. Two kilograms consumed with 83% conversion means 2 000 × 0.83 = 1 660 g embedded into polymer. Each molecule of the initiator initiates one chain, so the number-average molecular weight is 1 660 / 0.1471 = 11 288 g mol−1. After rounding to two digits, we get the number-average molecular weight of 11 000 g mol−1.
Poly(ε-caprolactone) can also be prepared by radical ring opening polymerization of 2-methylidene-1,3-dioxepane (C). [VISUAL]
How would you synthesize precursor C starting from butane-1,4-diol and bromoacetaldehyde dimethyl acetal (D)? Write the synthetic scheme.
Model Answer
[VISUAL] Synthesis scheme of precursor C from butane-1,4-diol and bromoacetaldehyde dimethyl acetal (D):
1. Transacetalization of D with butane-1,4-diol under acid catalysis (H+) to form 2-(bromomethyl-1,3-dioxepane.
2. Elimination of HBr using KOtBu to yield 2-methylidene-1,3-dioxepane (C).
Imagine dioxepane C was prepared from a 14C-labeled compound D (the labelled carbon is marked with an asterisk) and subjected to the radical polymerization reaction. [VISUAL]
Write the structure of poly(ε-caprolactone) and mark the radiolabelled carbon(s) with an asterisk.
Model Answer
[VISUAL] Poly(ε-caprolactone) structure where the carbonyl carbon is labeled with an asterisk: -[O-C*(=O-(CH2)5]-n
Proteins are natural polyamides based on α-amino acids. In living organisms, they are synthesized by translation based on genetic information, but they can also be prepared synthetically by a nucleophile-initiated ring opening polymerization. In this case, the activated cyclic monomers, N-carboxyanhydrides E (also called Leuchs' anhydrides) are used. They can be prepared by the reaction of an α-amino acid with phosgene:
[VISUAL]
Draw the structure of the activated monomer E formed from α-alanine (2-aminopropanoic acid).
Model Answer
[VISUAL] 4-methyloxazolidine-2,5-dione structure.
During polymerization, a gas is evolved and a polypeptide is formed.
Write the formula of the gas and the structure of the polymer formed from monomer E with butane-1-amine as initiator.
Model Answer
[VISUAL] Gas evolved: CO2
Structure of the polymer: H-[NH-CH(CH3)-CO]n-NH-(CH2)3-CH3
Natural proteins are formed exclusively from homochiral amino acids, i.e., only one enantiomer is present in the protein. This is vital for its 3D structure and function. Theoretically, if only a single amino acid in an enzyme is exchanged for its enantiomer, the chain changes its conformation, resulting in compromised catalytic efficiency.
Let us investigate lysozyme, a bacterial cell wall-lysing enzyme present in egg whites and tears. It contains 129 amino acid residues, 12 of which are glycines.
What would be the % yield of functional lysozyme if the proteosynthetic apparatus of the cell did not distinguish between enantiomers of the amino acids and had both enantiomers of amino acids available in equal quantities? Consider only the chirality on the α-carbon of all amino acids as the configuration on other chiral centres (in threonine and isoleucine) has only marginal effect on overall protein 3D structure. Note that only the enzyme digesting bacterial cell walls is claimed as functional.
Model Answer
A single wrong enantiomer of an amino acid in the protein structure causes loss of activity. Glycine is not chiral, so there are 129 − 12 = 117 chiral amino acids in lysozyme. The overall yield is (1/2)^117 × 100% = 6.02 × 10^−34 %.
Theoretically, in the 'world behind the mirror' the all-D-protein would be active against the all-chiral reversed proteoglycan. However, this does not meet the condition that only the enzyme digesting native peptidoglycan is considered functional.
In one egg there is ca 120 mg of lysozyme. How much protein (in kg) would you have to synthesize under the conditions described in 26.9 to produce enough functional lysozyme for one egg? Compare your result with the mass of the planet Earth (5.972 × 10^24 kg).
Model Answer
The amount of enzyme (120 mg = 0.000 12 kg) obtained with 6.02 × 10^−34 % yield (see the answer in 26.9) would require the production of 0.00012 / ((1/2)^117) = 1.99 × 10^31 kg of material. As the Earth weighs 5.972 × 10^24 kg, this corresponds to 1.99 × 10^31 / 5.972 × 10^24 = 3.34 × 10^6 times the mass of the Earth.