10 Marking scheme: End-of-chapter test

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1 1 a Element Carbon Hydrogen Oxygen Percentage Mass in 100 g 62.1 g 10.3 g 27.6 g Number of moles = 5.18 = 10.3 = Relative number of atoms = 3 = 6 = 1 empirical formula = C 3 H 6 O b molecular mass is 58 (from the m/e value for the penultimate peak). empirical formula mass = molecular mass, therefore the molecular formula = C 3 H 6 O c i ii It is propanal. iii There are three peaks present in the 1 H NMR spectrum; propan-2-one has only one type of proton and therefore it would have only one peak. d Three types of proton and therefore three peaks. The peak for the CH 3 protons at δ = 1.1 ppm is split into a triplet because of the two chemically different protons on the adjacent carbon. The peak for the CH 2 protons at δ = 2.5 ppm is split into a quadruplet because of the three chemically different protons on the adjacent carbon. The peak for the CHO proton at δ = 9.8 ppm is split into a triplet because of the two chemically different protons on the adjacent carbon. e The strong absorption at 1750 cm 1 which is due to the presence of the >C=O group. COAS Chemistry 2 Teacher Resources Original material Cambridge University Press 2005, 2007,

2 2 a Element Carbon Hydrogen Percentage Mass in 100g 90.6 g 9.4 g Number of moles in 100 g Relative number of atoms = 7.55 = = 1 = Whole numbers 4 5 empirical formula = C 4 H 5 b The m/e value for the penultimate peak on the mass spectrum = 106 this is twice the empirical formula mass, so the molecular formula is C 8 H 10 c i ii 1 mark for each structure [4] This is because when chlorinated it forms only the following compound: The other possible isomers form several different compounds. The 1 H NMR spectrum shows two peaks as C has only two different types of proton; the other possible isomers have larger numbers of different types of proton. d At δ = 7 ppm there is the peak from the benzene ring protons; at δ = 2.3 ppm the peak corresponds to the CH 3 protons. COAS Chemistry 2 Teacher Resources Original material Cambridge University Press 2005, 2007,

3 3 a Element Carbon Hydrogen Oxygen Percentage Mass in 100 g 77.8 g 7.41 g 14.8 g Number of moles in 100g = 6.48 = 7.41 = Relative number of atoms = 7 = 8 = 1 empirical formula = C 7 H 8 O b There is a strong absorption at ~3300 cm 1 corresponding to the OH group in alcohols. c The m/e value for the molecular-ion peak is 108; this corresponds to the mass of the empirical formula, so molecular formula = empirical formula = C 7 H 8 O d i 1 mark for each structure [5] ii iii The relative area of the peak at δ = 7.3 ppm is 5, so there are five protons on the benzene ring; therefore D cannot be a di-substituted compound. The other mono-substituted compound has only two types of proton while the 1 H NMR spectrum of D has three peaks, corresponding to the three different types of proton in D. e There are three peaks because of the three chemically different types of proton in D: the peak for five protons at δ = 7.3 ppm is caused by the five benzene protons; the peak given by two protons at δ = 4.6 ppm is caused by the CH 2 protons; the peak at δ = 2.4 ppm is caused by the OH proton. COAS Chemistry 2 Teacher Resources Original material Cambridge University Press 2005, 2007,

4 4 a Element Carbon Hydrogen Oxygen Percentage Mass in 100 g 69.8 g 11.6 g 18.6 g Number of moles in 100 g Relative number of atoms = 5.82 = 11.6 = = 5 = 10 = empirical formula = C 5 H 10 O b The m/e value for the molecular-ion peak is 86 so its relative molecular mass is 86; this is the same as the empirical formula mass, so the molecular formula is also C 5 H 10 O. c 1 mark for each structure 1 mark for each structure [7] d It must have a carbonyl group, because of the positive result with 2,4-DNPH. It is an aldehyde, because of the positive result from the silver mirror test. On the NMR spectrum there is an absorption at δ = 9.5 ppm, characteristic of the aldehyde CHO proton, and on the IR spectrum there is an absorption at 1720 cm 1, characteristic of a C=O group. e On the 1 H NMR spectrum there is an absorption at δ = 9.5 ppm, characteristic of the aldehyde CHO proton; it has only two distinct peaks and therefore only two types of proton; therefore E must be There is no splitting of the peaks because there are no adjacent carbons with chemically different protons on them. COAS Chemistry 2 Teacher Resources Original material Cambridge University Press 2005, 2007,

5 5 a CH 3 CH 2 CH 2 COOH CH 3 CH(CH 3 )COOH CH 3 COOCH 2 CH 3 CH 3 CH 2 COOCH 3 HCOOCH 2 CH 2 CH 3 HCOOCH(CH 3 ) 2 b F is butanoic acid, CH 3 CH 2 CH 2 COOH because it gives effervescence with Na 2 CO 3 and has four types of carbon atom. G is 2-methylpropanoic acid, CH 3 CH(CH 3 )COOH because it gives effervescence with Na 2 CO 3 and has three types of carbon atom. H is 2-methylethyl methanoate, HCOOCH(CH 3 ) 2 because it does not give effervescence with Na 2 CO 3 and has three types of carbon atom. COAS Chemistry 2 Teacher Resources Original material Cambridge University Press 2005, 2007,