22. Analytical techniques
A section of Chemistry, 9701
Listing 10 of 112 questions
When answering this question it should be assumed that together all the hydrogen atoms in a benzene ring result in a single unsplit peak at δ = 7.2 in a proton (1H) NMR spectrum. The structures of five isomeric ketones, P, Q, R, S and T are given. P C6H5COCH(CH3)2 S C6H5CH2CH2COCH3 Q C6H5COCH2CH2CH3 T C6H5CH(CH3)COCH3 R C6H5CH2COCH2CH3 Identify all the chiral carbon atoms on the structures above. Label each chiral carbon atom with an asterisk (*). The proton (1H) NMR spectrum of one of the five isomers, P, Q, R, S or T, is shown in Fig.8.1. / ppm Identify which of the compounds P, Q, R, S or T gives this spectrum. Draw the displayed formula of the compound you have identified. Identify the protons responsible for the peaks at δ = 3.7, δ = 2.5 and δ = 1.0 on the structure you have drawn.  Name the splitting pattern of the peak at δ = 3.7. Explain why it has this splitting pattern. Choose from the letters P, Q, R, S and T to identify: the two compounds that each have a doublet peak in the proton (1H) NMR spectrum the compound with only three peaks in its proton (1H) NMR spectrum. Suggest a suitable solvent that should be used for obtaining the spectrum shown in Fig.8.1. The proton (1H) NMR spectrum of compoundT is compared in the presence of D2O and in the absence of D2O. Describe any difference between the two spectra. Explain your answer. Complete Table8.1 below to give the number of peaks in the carbon-13 NMR spectrum of each compound. Table 8.1 compound number of peaks compound number of peaks  
9701_w22_qp_42
THEORY
2022
Paper 4, Variant 2
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Gas-liquid chromatography involves a stationary phase and a mobile phase. Name, or describe in detail, a suitable substance that could be used for each phase. stationary mobile  A mixture of three organic compounds is separated by gas-liquid chromatography. The chromatogram obtained is shown in Fig.9.1. The amount of each substance is proportional to the area under its peak. retention time recorder response A C B Explain the meaning of retention time. Calculate the percentage of B in the mixture. Show your working.  percentage of B = % Complete Table9.1 to give the number of peaks in the carbon-13 NMR spectrum of each of the five isomers of C5H10O2 that has an ester group. Table 9.1 structural formula number of peaks CH3CH2CH2CO2CH3 CH3CH2CO2CH2CH3 CH3CO2CH2CH2CH3 (CH3)2CHCO2CH3 CH3CO2CH(CH3)2  State the number of peaks that would be seen in the proton (1H) NMR spectrum of methylbutanoate, CH3CH2CH2CO2CH3. Name all the splitting patterns seen in this spectrum. number of peaks splitting patterns  D and E are both esters with the molecular formula C5H10O2. Their proton (1H) NMR spectra are shown in Fig.9.2 and Fig.9.3. chemical shift, D chemical shift, E Table 9.2 environment of proton example typical 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–O, –CH2–O, –CH2–Cl 3.2–4.0 attached to alkene =CHR 4.5–6.0 Deduce the structures of the two esters D and E and draw their displayed formulae in the boxes below. D C5H10O2 E C5H10O2  The spectrum of D includes a quartet at δ4.1. Identify the protons responsible for this quartet on your structure in by labelling these protons with the letter F. Explain why this peak is split into a quartet.  The spectrum of E has a doublet at δ1.1. Identify the protons responsible for this doublet on your structure in by labelling these protons with the letter G. Explain why this peak has a chemical shift of 1.1.  
9701_w22_qp_43
THEORY
2022
Paper 4, Variant 3
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Describe what is meant by a racemic mixture. Asparagine is an amino acid that contains a chiral carbon atom and displays stereoisomerism. Separate samples of asparagine are dissolved in CDCl 3 and analysed using carbon‑13 and proton (1H) NMR spectroscopy. C C asparagine H2N CH2 NH2 H O OH O C 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  The isoelectric point of asparagine, asn, is at pH5.4. Describe the meaning of the term isoelectric point. Draw the structure of asparagine at pH1.0.  Asparagine can polymerise to form poly. Draw the structure of poly, showing two repeat units. The peptide linkage should be shown displayed.  The isoelectric point of lysine, lys, is at pH9.8. H2N C lysine H (CH2)4 NH2 COOH 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. mixture applied here E F G + – Suggest identities for the species responsible for spots E, F and G. Explain your answers. spot identity E F G  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 Table6.1 with an example of each of these. Table 6.1 stationary phase mobile phase thin-layer chromatography gas-liquid chromatography  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. P unknown amino acid solvent 1 Q R S P unknown amino acid solvent 2 Q R S cm Identify the unknown amino acid. Justify your answer. A mixture containing three organic compounds is analysed by gas chromatography and mass spectrometry. The gas chromatogram is shown. peak area / mm2 J J K K L L retention time / minutes 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. [Mr: L, 116]  concentration of L = mol dm–3 
9701_s22_qp_41
THEORY
2022
Paper 4, Variant 1
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Describe what is meant by a racemic mixture. Asparagine is an amino acid that contains a chiral carbon atom and displays stereoisomerism. Separate samples of asparagine are dissolved in CDCl 3 and analysed using carbon‑13 and proton (1H) NMR spectroscopy. C C asparagine H2N CH2 NH2 H O OH O C 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  The isoelectric point of asparagine, asn, is at pH5.4. Describe the meaning of the term isoelectric point. Draw the structure of asparagine at pH1.0.  Asparagine can polymerise to form poly. Draw the structure of poly, showing two repeat units. The peptide linkage should be shown displayed.  The isoelectric point of lysine, lys, is at pH9.8. H2N C lysine H (CH2)4 NH2 COOH 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. mixture applied here E F G + – Suggest identities for the species responsible for spots E, F and G. Explain your answers. spot identity E F G  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 Table6.1 with an example of each of these. Table 6.1 stationary phase mobile phase thin-layer chromatography gas-liquid chromatography  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. P unknown amino acid solvent 1 Q R S P unknown amino acid solvent 2 Q R S cm Identify the unknown amino acid. Justify your answer. A mixture containing three organic compounds is analysed by gas chromatography and mass spectrometry. The gas chromatogram is shown. peak area / mm2 J J K K L L retention time / minutes 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. [Mr: L, 116]  concentration of L = mol dm–3 
9701_s22_qp_43
THEORY
2022
Paper 4, Variant 3
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