A-LevelChemistryIntroduction to organic chemistry9701/41 · May/June 2022 · Q612 Marks

Asparagine is an amino acid that contains a chiral carbon atom and displays stereoisomerism. Separate samples of asparagine are dissolved in CDCl₃ and analysed using carbon-13 and proton (¹H) NMR spectroscopy. [Figure 6.1 showing asparagine structure]

(a)Describe what is meant by a racemic mixture.[1]
(b)Predict the number of peaks seen in the carbon-13 and proton (¹H) NMR spectra of asparagine. carbon-13 NMR proton (¹H) NMR number of peaks[1]
(c)(i)The isoelectric point of asparagine, asn, is at pH 5.4. Describe the meaning of the term isoelectric point.[1]
(c)(ii)Draw the structure of asparagine at pH 1.0.[1]
(d)Asparagine can polymerise to form poly(asparagine). Draw the structure of poly(asparagine), showing two repeat units. The peptide linkage should be shown displayed.[2]
(e)The isoelectric point of lysine, lys, is at pH 9.8. [Figure 6.2 showing lysine structure] A mixture of the dipeptide lys-asn and its two constituent amino acids, asparagine and lysine, is analysed by electrophoresis using a buffer at pH 5.0. The results obtained are shown in Fig. 6.3. [Figure 6.3 showing electrophoresis results] Suggest identities for the species responsible for spots E, F and G. Explain your answers. spot identity E F G[3]
(f)(i)Thin-layer and gas-liquid chromatography can be used to analyse mixtures of substances. Each type of chromatography makes use of a stationary phase and a mobile phase. Complete Table 6.1 with an example of each of these. Table 6.1 stationary phase mobile phase thin-layer chromatography gas-liquid chromatography[1]
(f)(ii)An unknown amino acid is analysed using thin-layer chromatography. Two chromatographs of the unknown amino acid and four reference amino acids, P, Q, R and S, are obtained using two different solvents. [Figure 6.4 showing two chromatograms] Identify the unknown amino acid. Justify your answer.[1]
(g)A mixture containing three organic compounds is analysed by gas chromatography and mass spectrometry. The gas chromatogram is shown. [Figure 6.5 showing gas chromatogram and table of peak areas] The area underneath each peak is proportional to the mass of the respective compound in the mixture. The concentration of K in the mixture is 5.52 × 10⁻² g dm⁻³. Calculate the concentration, in mol dm⁻³, of compound L in the mixture. [M_r: L, 116] concentration of L =[1]

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Mark scheme

Marking points from the official mark scheme (May/June 2022). The mark or mark type is shown in brackets.

[1] a mixture containing equal amounts of each optical isomer
[1] carbon-13 NMR: 4, proton NMR: 5
[1] the pH at which an amino acid exists as a zwitterion
OR the pH at which an amino acid has no overall charge
[1] (Structure as shown in MS: H₂N-C(=O)-CH₂-CH(NH₃⁺)-C(=O)OH)
[2] M1 peptide bond displayed
M2 rest of the structure correct + continuation bonds
[3] M1 table correctly completed (E: Asn, F: Lys-Asn, G: Lys)
M2 Lys and Lys-Asn are positively charged OR Asn is (nearly) uncharged
M3 Lys-Asn has the highest Mr
[1] aluminum oxide / silica (on solid support) AND inert gas / named inert gas e.g. N2
[1] S AND Rf is the same as the unknown amino acid in both solvents
[1] mass of L = 58 / 44 * 5.52 × 10⁻² = 7.28 × 10⁻² g
conc. of L = 7.28 × 10⁻²/ 116 = 6.27 × 10⁻⁴ (mol dm⁻³) min 2sf