Lactic acid, CH3CH(OH)COOH, is the only monomer needed to form the polymer polylactic acid, PLA.
- (a)(i)Draw a short length of the PLA polymer chain, including a minimum of two monomer residues. The methyl groups may be written as –CH3 but all other bonds should be shown fully displayed. Label one repeat unit of polylactic acid on your diagram.[2]
- (a)(ii)Give the name of the type of polymerisation involved in the formation of PLA and the name of the functional group that forms between the monomers. type of polymerisation .......................................................................................................... functional group ..........................................................................................................[1]
- (a)(iii)Predict whether PLA is readily biodegradable. Explain your answer.[1]
- (b)(i)The proton (1H) NMR spectrum of CH3CH(OH)COOH in CDCl3 is shown in Fig. 8.1. The proton NMR chemical shift ranges are shown in Table 8.1. Lactic acid [Figure 8.1 showing NMR spectrum] Table 8.1 environment of proton example chemical shift range δ/ppm alkane -CH3, -CH2-, >CH– 0.9-1.7 alkyl next to C=O CH3-C=O, -CH2–C=O, >CH–C=O 2.2-3.0 alkyl next to aromatic ring CH3-Ar, -CH2-Ar, >CH-Ar 2.3-3.0 alkyl next to electronegative atom CH3-0, -CH2-O, –CH2–Cl 3.2-4.0 attached to alkene =CHR 4.5-6.0 attached to aromatic ring H-Ar 6.0-9.0 aldehyde HCOR 9.3-10.5 alcohol ROH 0.5-6.0 phenol Ar-OH 4.5-7.0 carboxylic acid RCOOH 9.0-13.0 Use Fig. 8.1 and Table 8.1 to complete Table 8.2. Table 8.2 proton environment chemical shift (δ) name of splitting pattern -COOH >CH -OH -CH3[3]
- (b)(ii)Name the substance responsible for the peak at δ = 0.0.[1]
- (b)(iii)Explain why CDCl3 is a better solvent than CHCl3 for use in proton NMR.[1]
- (c)(i)An impure sample of CH3CH(OH)COOH contains pentan-3-one as the only contaminant. The mixture is analysed using gas/liquid chromatography. The pentan-3-one is found to have a longer retention time than the lactic acid. Explain what is meant by retention time.[1]
- (c)(ii)Suggest suitable substances, or types of substances, that could be used as the mobile and stationary phases. mobile phase .......................................................................................................... stationary phase ..........................................................................................................[1]
- (c)(iii)Describe how the percentage composition of the mixture can be determined from the gas/liquid chromatogram.[1]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2023). The mark or mark type is shown in brackets.
- [1] (a)(i) diagram has trailing bonds, two or more monomer residues AND either repeat unit correct OR ester linkage correct
- [1] (a)(i) diagram fully correct
- [1] (a)(ii) condensation AND ester
- [1] (a)(iii) it is biodegradable because it can hydrolyse
- [1] (b)(i) d values: 12, 4.4
- [1] (b)(i) 5.0, 1.6
- [1] (b)(i) splitting patterns: singlet, quartet / quadruplet, singlet, doublet
- [1] (b)(ii) TMS/tetramethylsilane
- [1] (b)(iii) CDC13 will not give an absorption / peak OR CHCl3 will give an absorption / peak
- [1] (c)(i) time between injection and detection
- [1] (c)(ii) an unreactive gas AND a non-polar liquid
- [1] (c)(iii) area of peak divided by total area of all peaks × 100%
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