A-LevelChemistryPolymerisation9701/42 · Feb/Mar 2025 · Q614 Marks
Maleic anhydride is an unsaturated cyclic compound used in the formation of several polymers. Maleic anhydride can be used to form maleic acid and tartaric acid. [Figure 6.1] Maleic acid reacts with ethane-1,2-diol to form a condensation polymer. Draw a section of this polymer, showing only one repeat unit. The new functional group formed should be shown fully displayed.
- (a)(i)Maleic anhydride is an unsaturated cyclic compound used in the formation of several polymers. Maleic anhydride can be used to form maleic acid and tartaric acid. [Figure 6.1] Maleic acid reacts with ethane-1,2-diol to form a condensation polymer. Draw a section of this polymer, showing only one repeat unit. The new functional group formed should be shown fully displayed.[2]
- (a)(ii)Identify a suitable reagent and the conditions for reaction 2.[1]
- (b)(i)Compound U can be formed from tartaric acid in two steps, as shown in Fig. 6.2. [Figure 6.2] Suggest the type of reaction that occurs in step 1.[1]
- (b)(ii)U can act as a bidentate ligand. Explain what is meant by a bidentate ligand.[2]
- (b)(iii)Complete the three-dimensional diagrams in Fig. 6.3 to show both stereoisomers of [Cu(U)3]4-. Use O O to represent ligand U. [Figure 6.3][2]
- (c)(i)A student analyses an aromatic compound, X, C8H8O3, using NMR spectroscopy. Fig. 6.4 shows the carbon-13 NMR spectrum of a sample of X dissolved in D2O. [Figure 6.4] Separate samples of X were analysed using proton (1H) NMR spectroscopy. Table 6.1 gives information obtained from this analysis. Table 6.1 solvent number of signals in proton (1H) NMR spectrum CDC13 6 D2O 4 Identify the number of different carbon environments present in X.[1]
- (c)(ii)Explain why X is dissolved in D₂O before obtaining its proton (1H) NMR spectrum.[1]
- (c)(iii)Aromatic compound X gives a yellow precipitate when it reacts with alkaline I2(aq). Use the information in (c) to suggest a structure for X. Explain your reasoning. X, C8H8O3 [Table 6.2][4]
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Mark scheme
Marking points from the official mark scheme (Feb/Mar 2025). The mark or mark type is shown in brackets.
- [2] M1: ester group in middle displayed between two monomers
- M2: rest of molecule correct (only one repeat unit)
- [1] cold acidified dilute KMnO4/MnO4-
- [1] condensation OR dehydration OR elimination
- [2] species with two lone pairs / LP of electrons)
- that form dative (covalent) / co-ordinate bond(s) to a (central) transition element / metal atom/ion
- [2] M1 one correct diagram
- M2 correct diagram of the other optical isomer
- [1] 8 / eight
- [1] to remove peaks / signals / absorptions from acidic protons / OH / COOH / labile protons
- [4] Molecule must not be symmetrical (as eight different C environments)
- Viable structures of X
- methyl ketone / CH3CO group because (yellow) ppt / CHI3 formed (with alkaline I2(aq))
- any two linked statements: two -OH groups because 1H NMR signals lost in D2O
- ketone / carbonyl / CH3CO as (13C NMR) signal with Δ = 205 / 190–220 ppm
- methyl / carbon next to C=O because (13C NMR) signal with Δ = 30–65 ppm
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