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A-Level · Chemistry

AC8 Analytical Techniques

AQA 7405 · Calculator allowed · about 150 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

A student analyses a sample of 2-chlorobutane, C4H9Cl, using a mass spectrometer.

(a)State what is meant by the term 'molecular ion' in mass spectrometry.(1)

(b)The mass spectrum of 2-chlorobutane shows the molecular ion split into two peaks, at m/z = 92 and m/z = 94, with relative peak heights in the ratio 3:1. Explain the origin of these two peaks and explain why they occur in this ratio.(3)

(c)Calculate the relative molecular mass, Mr, of 2-chlorobutane. Use Ar: C = 12.0, H = 1.0, Cl = 35.5. Show your working.(2)

(d)A fragment ion peak is seen at m/z = 57. This fragment forms by loss of a chlorine radical, Cl., from the molecular ion at m/z = 92. Deduce the molecular formula of this fragment and suggest a structure for it.(3)

(Total for Question 1 is 9 marks)

Infrared (IR) spectroscopy can be used to identify functional groups. Typical absorption ranges: O-H (carboxylic acid) 2500-3300 cm-1 (broad); O-H (alcohol) 3230-3550 cm-1 (broad); N-H 3300-3500 cm-1; C=O 1680-1750 cm-1; C-O 1000-1300 cm-1; C-H 2850-3100 cm-1.

(a)The infrared spectra of ethanoic acid, CH3COOH, and propanone, CH3COCH3, both show an absorption at about 1715 cm-1. Identify the bond responsible for this absorption.(1)

(b)The infrared spectrum of ethanoic acid shows a broad absorption between 2500 and 3300 cm-1 that is absent from the spectrum of propanone. Explain how this feature allows the two compounds to be distinguished.(2)

(c)A sample of car exhaust gas strongly absorbs infrared radiation at 2349 cm-1, characteristic of the C=O bond vibration in carbon dioxide. Explain, in terms of bond behaviour, why CO2 absorbs infrared radiation but the diatomic gas N2 does not, and state the environmental significance of this property.(3)

(d)An unknown liquid, X, gives an IR spectrum with a strong, sharp absorption at 1735 cm-1 and no absorption between 2500 and 3550 cm-1. Suggest a functional group that is present in X and one that must be absent.(2)

(Total for Question 2 is 8 marks)

Required practical: A student separates a mixture of three amino acids, alanine, glycine and leucine, using thin-layer chromatography (TLC), following the required practical procedure for separating species by TLC and interpreting Rf values.

(a)State two precautions that should be taken when spotting the amino acid mixture onto the TLC plate, to obtain reliable results.(2)

(b)The solvent front travelled 8.4 cm from the baseline. The spot corresponding to glycine travelled 3.2 cm. Calculate the Rf value for glycine.(2)

(c)Amino acids are colourless, so the plate must be treated with a locating agent to make the spots visible. Name a suitable locating agent for amino acids and state the colour change it produces.(2)

(d)Reference standards run under the same conditions give Rf values: alanine 0.51, glycine 0.38, leucine 0.74. Using the table and your answer to part (b), identify which amino acid the unknown spot in part (b) is, and explain why comparing Rf values under identical conditions is essential for reliable identification.(3)

(e)Suggest one reason why paper chromatography might give poorer separation than TLC for amino acids with very similar Rf values, and suggest one modification that could improve the separation of such closely running compounds.(2)

(f)A repeat run gives an Rf value of 0.35 for the same amino acid (glycine), compared with 0.38 previously. Suggest two reasons for this small variation between repeats.(2)

(g)State why the TLC plate should not be handled with bare fingers before or during the experiment.(1)

(h)A student suggests using UV light instead of ninhydrin to visualise the amino acid spots. Evaluate whether this is likely to work, giving a reason.(2)

(i)State why it is important to use the same TLC plate (or plates from the same batch) when comparing the Rf value of an unknown with reference standards, rather than plates from a different manufacturer or batch.(2)

(Total for Question 3 is 18 marks)

Three isomers of molecular formula C3H8O are propan-1-ol (CH3CH2CH2OH), propan-2-ol ((CH3)2CHOH) and methoxyethane (CH3OCH2CH3).

(a)State what is meant by a 'carbon environment' in 13C NMR spectroscopy.(1)

(b)Deduce the number of peaks expected in the 13C NMR spectrum of each of the three isomers.(3)

(c)A sample is analysed by 13C NMR and gives exactly two peaks. Identify which isomer this is, and use the approximate shift ranges C-C (alkyl) 0-50 ppm and C-O 50-90 ppm to justify the expected shift of each peak.(3)

(d)Explain why 13C NMR spectroscopy alone cannot always distinguish between structural isomers with the same number of carbon environments, and state one other technique that could be used to confirm the identity of an unknown sample.(2)

(Total for Question 4 is 9 marks)

The 1H NMR spectrum of propanoic acid, CH3CH2COOH, is recorded.

(a)State the 'n+1 rule' used to predict the multiplicity (splitting pattern) of a proton signal in 1H NMR spectroscopy.(1)

(b)Predict the splitting pattern and relative integration for each of the three proton environments in propanoic acid.(4)

(c)The COOH proton in propanoic acid appears at about 11.8 ppm. When a small amount of D2O is added to the sample and the spectrum re-run, this peak disappears. Explain this observation.(2)

(d)Explain why tetramethylsilane (TMS) is used as the standard reference compound in 1H NMR spectroscopy.(2)

(Total for Question 5 is 9 marks)

Two isomers with molecular formula C3H6O are propanal, CH3CH2CHO, and propanone, CH3COCH3.

(a)Show that the relative molecular mass of C3H6O is 58. Use Ar: C = 12.0, H = 1.0, O = 16.0.(2)

(b)The mass spectrum of propanone shows a major fragment peak at m/z = 43, formed by loss of a neutral fragment of mass 15 from the molecular ion (58). Identify the fragment lost and the ion formed, including its formula.(2)

(c)The mass spectrum of propanal shows a major fragment peak at m/z = 29, formed by loss of a fragment of mass 29 from the molecular ion (58). Suggest the identity of the ion at m/z = 29 and explain how this fragment could help distinguish propanal from propanone.(3)

(d)State one other analytical technique, other than mass spectrometry, that could distinguish propanal from propanone, and identify the key spectral difference that would be observed.(2)

(Total for Question 6 is 9 marks)

A forensic chemist uses gas chromatography (GC) to determine the concentration of ethanol in a blood sample, using a calibration curve produced from ethanol standards of known concentration.

(a)State one advantage of gas chromatography over thin-layer chromatography for this type of quantitative analysis.(1)

(b)The retention time of ethanol under these GC conditions is 2.3 minutes, and a second peak for propan-1-ol (an internal standard) appears at 3.6 minutes. Explain what is meant by 'retention time' and why it can be used to help identify a compound.(2)

(c)The calibration line for ethanol is: peak area = 42.5c + 15, where c is the concentration of ethanol in mg per 100 cm3 of blood. A blood sample gives a peak area of 780. Calculate the concentration of ethanol, in mg per 100 cm3, in the blood sample.(3)

(d)The legal drink-drive limit in England is 80 mg of alcohol per 100 cm3 of blood. Using your answer to part (c), state whether the sample is above or below the legal limit, and calculate the percentage of the legal limit that this concentration represents.(2)

(e)Explain why using an internal standard, such as propan-1-ol, improves the reliability of quantitative GC analysis.(2)

(Total for Question 7 is 10 marks)

An unknown organic compound, Y, has molecular formula C4H8O2 (Mr = 88). Mass spectrum: molecular ion peak at m/z = 88; fragment peaks at m/z = 43 and m/z = 45. Infrared spectrum: strong absorption at 1740 cm-1; no absorption between 2500 and 3300 cm-1. 1H NMR spectrum: singlet at 2.0 ppm (3H); quartet at 4.1 ppm (2H); triplet at 1.3 ppm (3H).

(a)Use the infrared data to deduce the type of carbonyl-containing functional group present in Y, giving a reason.(2)

(b)Use the 1H NMR data to deduce the number of different proton environments in Y and calculate the total number of hydrogen atoms shown by the integration values.(2)

(c)The fragment ion at m/z = 43 is formed by loss of a group of mass 45 from the molecular ion (88). Suggest the identity of the group lost and the fragment ion formed.(2)

(d)Using all the data given, deduce the structural formula of Y and explain how the 1H NMR splitting patterns support your answer.(4)

(Total for Question 8 is 10 marks)

A brominated organic compound gives a mass spectrum with a molecular ion cluster of two peaks of approximately equal height at m/z = 122 and m/z = 124.

(a)Explain why bromine-containing compounds give two molecular ion peaks of approximately equal (1:1) height, two mass units apart.(2)

(b)A different compound contains two chlorine atoms in its structure. State the ratio of peak heights expected for the M : M+2 : M+4 peaks in its mass spectrum, and explain your reasoning.(3)

(c)Propan-1-ol (C3H8O) and ethanoic acid (C2H4O2) both have a nominal (integer) relative molecular mass of 60. Using exact atomic masses C = 12.0000, H = 1.0078, O = 15.9949, calculate the exact mass of each compound to 3 decimal places, and explain how high resolution mass spectrometry could distinguish between them despite their equal nominal Mr.(5)

(Total for Question 9 is 10 marks)

Methyl propanoate, CH3CH2COOCH3, is a structural isomer of ethyl ethanoate (see Q8), both with molecular formula C4H8O2.

(a)Predict the number of proton environments in methyl propanoate and give the expected splitting pattern for each.(4)

(b)Explain how the 1H NMR spectrum of methyl propanoate could be distinguished from that of its isomer, ethyl ethanoate (Q8), even though both show a singlet, a quartet and a triplet.(3)

(c)A chemist obtains a 13C NMR spectrum of an unknown C4H8O2 ester and finds four distinct peaks. Explain why this result alone cannot be used to distinguish between methyl propanoate and ethyl ethanoate, and state what further data would allow the isomers to be distinguished.(2)

(Total for Question 10 is 9 marks)

An unknown organic compound, Z, has molecular formula C4H8O (Mr = 72). Data obtained: Mass spectrum: molecular ion peak at m/z = 72; fragment peaks at m/z = 57 and m/z = 29. Infrared spectrum: strong absorption at 1715 cm-1; no absorption between 2500 and 3550 cm-1. 1H NMR spectrum: doublet at 1.15 ppm (6H); multiplet at 2.55 ppm (1H); doublet at 9.60 ppm (1H). Using all the data provided, deduce the structural formula of Z. Your answer should use evidence from the mass spectrum, the infrared spectrum and the 1H NMR spectrum to justify the functional group present, the carbon skeleton, and the final structure.

(Total for Question 11 is 6 marks)

An anti-doping laboratory tests an athlete's urine sample for a banned substance, X, using GC-MS with an internal standard, IS, added to the sample before extraction. The peak area of X is 5400 and the peak area of IS is 3600. Calibration established the relationship (Area X / Area IS) = k x (mass X / mass IS), with k = 1.20. A 2.00 cm3 urine sample has 50.0 ng of internal standard added before extraction.

(a)Calculate the peak area ratio (Area X : Area IS) for this sample.(1)

(b)Using (Area X / Area IS) = k x (mass X / mass IS), with k = 1.20 and mass IS = 50.0 ng, calculate the mass of X, in ng, present in the 2.00 cm3 sample.(3)

(c)Calculate the concentration of X in the urine sample, in ng per cm3.(2)

(d)The World Anti-Doping Agency reporting threshold for X is 30 ng per cm3. State, with a reason, whether this sample would be reported as a positive result, and suggest why laboratories typically require a second, independent confirmatory test (e.g. using a different column or ionisation method) before reporting an adverse finding.(3)

(e)Explain why using an internal standard, rather than relying on the absolute peak area of X alone, increases confidence in the accuracy of this quantitative result.(2)

(Total for Question 12 is 11 marks)

Mark scheme · AC8 Analytical Techniques

Question 1

(a) B1 the ion (peak) formed when a molecule loses an electron without fragmenting, giving a peak that represents the relative molecular mass (Mr) of the compound oe
(a) Answer: The ion formed when a molecule loses one electron without breaking apart; its m/z value gives the Mr of the compound.
(b) B1 chlorine exists naturally as two isotopes, 35Cl and 37Cl oe
(b) B1 the peak at m/z = 92 is the molecular ion containing 35Cl and the peak at m/z = 94 is the molecular ion containing 37Cl
(b) B1 the natural abundance of 35Cl to 37Cl is approximately 3:1 (accept approx. 75%:25%), so the peak heights reflect this abundance ratio
(b) Answer: Two peaks arise from the two chlorine isotopes (35Cl, 37Cl); their natural abundance ratio is about 3:1, matching the peak height ratio.
(c) M1 correct substitution: 4(12.0) + 9(1.0) + 35.5
(c) A1 92.5 (cao)
(c) Answer: Mr = 92.5
(d) M1 fragment formula is C4H9+
(d) A1 mass check: 4(12) + 9(1) = 57, consistent with m/z = 57 (cao)
(d) B1 suggests a valid carbocation structure, e.g. CH3CH2CH2CH2+ or CH3CH2+CHCH3 (any structure with formula C4H9+ accepted)
(d) Answer: C4H9+, e.g. the butyl cation CH3CH2CH2CH2+ (or the isomeric sec-butyl cation).

Question 2

(a) B1 C=O (carbonyl) bond
(a) Answer: C=O bond.
(b) B1 the broad absorption at 2500-3300 cm-1 is due to the O-H bond in the carboxylic acid (-COOH) group of ethanoic acid
(b) B1 propanone has no O-H bond (only a C=O group), so this absorption is absent, allowing the two compounds to be distinguished
(b) Answer: The broad O-H absorption (2500-3300 cm-1) is present in ethanoic acid but absent in propanone, which has no O-H bond.
(c) B1 bond vibration (stretching/bending) in CO2 causes a change in the dipole moment of the molecule oe
(c) B1 N2 is a symmetrical diatomic molecule with no permanent dipole, and its bond vibration causes no change in dipole moment, so it does not absorb IR radiation oe
(c) B1 because CO2 absorbs (and re-emits) infrared radiation, it acts as a greenhouse gas, contributing to the greenhouse effect/global warming
(c) Answer: CO2's bond vibrations change its dipole moment so it absorbs IR; symmetrical N2 has no dipole change on vibration so does not absorb IR. This makes CO2 a greenhouse gas.
(d) B1 present: a carbonyl-containing group such as an ester (or aldehyde/ketone), shown by the C=O absorption near 1735 cm-1
(d) B1 absent: O-H (no alcohol or carboxylic acid group present), as there is no absorption in the 2500-3550 cm-1 region
(d) Answer: Present: C=O (consistent with an ester); absent: O-H.

Question 3

(a) B1 use a fine capillary tube to apply a small, concentrated spot (allowing it to dry between applications)
(a) B1 draw the origin line in pencil (not ink) and keep the spot/baseline above the level of the solvent in the developing tank
(a) Answer: Use a fine capillary tube for a small, concentrated spot; keep the baseline above the solvent level and mark it in pencil.
(b) M1 Rf = distance travelled by spot / distance travelled by solvent = 3.2/8.4
(b) A1 0.38 (awrt 0.38)
(b) Answer: Rf = 0.38
(c) B1 ninhydrin (spray reagent)
(c) B1 the plate is sprayed and heated (e.g. in an oven or with a hairdryer), producing purple/pink spots
(c) Answer: Ninhydrin; sprayed on and heated to give purple/pink spots.
(d) B1 glycine (Rf = 0.38 matches the reference value)
(d) B1 Rf values depend on conditions such as the solvent used, temperature and the stationary phase, so the reference standards and the unknown must be run under identical conditions
(d) B1 running standards alongside the unknown on the same plate (rather than comparing with literature values) is more reliable, since exactly matching conditions are guaranteed
(d) Answer: Glycine; Rf is condition-dependent, so identical conditions (ideally the same plate) are needed for a valid comparison.
(e) B1 paper fibres are less uniform than the fine, even layer of stationary phase on a TLC plate, so paper chromatography generally gives lower resolution/more diffuse spots
(e) B1 modification: use two-dimensional chromatography (developing in a second solvent at right angles) or a different/more suitable solvent system, to better separate closely running spots
(e) Answer: TLC's uniform stationary phase gives better resolution than paper; two-dimensional chromatography (or a different solvent) can separate close-running spots further.
(f) B1 a slight difference in laboratory/tank temperature between the two runs (Rf is temperature dependent)
(f) B1 a slight variation in the exact solvent composition/volume used, or the tank not being fully saturated with solvent vapour (chamber not sealed/equilibrated) before development
(f) Answer: Small differences in temperature and/or solvent conditions (e.g. tank vapour saturation) between repeats.
(g) B1 oils/grease/sweat from fingers could contaminate the plate, producing extra spots or affecting Rf values and giving inaccurate/misleading results
(g) Answer: Skin oils can contaminate the plate and affect the spots/Rf values obtained.
(h) B1 UV visualisation only works for compounds that absorb/fluoresce under UV light (or plates containing a fluorescent indicator, showing spots as dark patches)
(h) B1 most amino acids (other than aromatic ones such as tryptophan, tyrosine or phenylalanine) do not absorb/fluoresce strongly under UV, so ninhydrin (a chemical locating agent) is more suitable/reliable for a general amino acid mixture
(h) Answer: UV light is unlikely to reveal all the spots reliably, since most amino acids do not absorb/fluoresce under UV; ninhydrin remains the more suitable locating agent.
(i) B1 different batches/manufacturers of TLC plate may have slightly different stationary phase thickness, particle size or activation, which can affect Rf values
(i) B1 using the same plate/batch (or running standards and the unknown side by side on the same plate) ensures conditions are identical, making the Rf comparison valid/reliable
(i) Answer: Different plate batches can give slightly different Rf values, so using the same plate/batch ensures a fair, valid comparison.

Question 4

(a) B1 carbon atoms that are in identical/equivalent chemical surroundings (bonded to the same groups, or related by symmetry), giving a single peak in the 13C NMR spectrum
(a) Answer: Carbon atoms in equivalent chemical surroundings, which give a single NMR peak.
(b) B1 propan-1-ol: 3 peaks (three different carbon environments, C1, C2, C3)
(b) B1 propan-2-ol: 2 peaks (two equivalent CH3 groups related by symmetry, plus one CH)
(b) B1 methoxyethane: 3 peaks (OCH3, CH2 and CH3 are all different environments)
(b) Answer: Propan-1-ol: 3 peaks; propan-2-ol: 2 peaks; methoxyethane: 3 peaks.
(c) B1 propan-2-ol (the only isomer that gives exactly two peaks)
(c) B1 the two equivalent CH3 carbons appear at a low shift, in the 0-50 ppm alkyl (C-C) range (approx. 25 ppm)
(c) B1 the CH carbon (bonded to OH) appears at a higher shift, in the 50-90 ppm (C-O) range (approx. 64 ppm)
(c) Answer: Propan-2-ol; CH3 carbons approx. 25 ppm (C-C range), CH carbon approx. 64 ppm (C-O range).
(d) B1 two different structures may coincidentally have the same number of (non-equivalent) carbon environments, giving the same number of peaks, so peak count alone is not conclusive
(d) B1 1H NMR (or IR, or mass spectrometry) can provide additional structural information, e.g. splitting patterns and integration, to confirm the structure
(d) Answer: Peak count alone can be ambiguous between isomers; 1H NMR (or IR/mass spectrometry) provides additional confirming evidence.

Question 5

(a) B1 a proton (or group of equivalent protons) with n equivalent neighbouring (adjacent) protons is split into (n+1) peaks/lines
(a) Answer: A proton with n equivalent neighbouring protons is split into (n+1) peaks.
(b) B1 CH3 protons: triplet, relative area 3 (split by 2 neighbouring CH2 protons, n=2, so n+1=3)
(b) B1 CH2 protons: quartet, relative area 2 (split by 3 neighbouring CH3 protons, n=3, so n+1=4)
(b) B1 COOH proton: singlet, relative area 1 (no adjacent protons on a neighbouring carbon to split it)
(b) B1 correct overall integration ratio 3:2:1 for CH3:CH2:COOH
(b) Answer: CH3: triplet (3H); CH2: quartet (2H); COOH: singlet (1H); ratio 3:2:1.
(c) B1 the O-H proton is exchangeable and is replaced by deuterium from D2O: -COOH + D2O -> -COOD + HOD oe
(c) B1 deuterium (D) does not give a signal in a 1H NMR spectrum, so the peak at 11.8 ppm disappears
(c) Answer: The exchangeable O-H proton is replaced by deuterium (forming O-D), which gives no signal in 1H NMR, so the peak vanishes.
(d) B1 TMS gives a single, strong absorption peak (all 12 H are equivalent) that does not overlap with other proton signals
(d) B1 TMS is unreactive, volatile and non-toxic and is easily removed from the sample, so it is used to fix the reference point of 0 ppm on the chemical shift scale
(d) Answer: TMS gives one sharp, non-overlapping peak and is inert/volatile/easy to remove, making it a convenient 0 ppm reference.

Question 6

(a) M1 correct substitution: 3(12.0) + 6(1.0) + 16.0
(a) A1 = 58 (cso, correct answer given, working must be shown)
(a) Answer: 58
(b) B1 fragment lost: a methyl radical, .CH3 (mass 15)
(b) B1 ion formed: CH3CO+ (acylium ion, formula C2H3O+, m/z = 43)
(b) Answer: .CH3 lost; CH3CO+ (C2H3O+) formed at m/z = 43.
(c) B1 ion at m/z = 29 is CHO+ (formyl/acylium cation), formed by loss of the ethyl radical, .C2H5
(c) B1 propanone's spectrum does not show a CHO+ (m/z = 29) fragment (its characteristic fragment is CH3CO+ at m/z = 43), so the presence or absence of m/z = 29 distinguishes the two isomers
(c) B1 both compounds show the same molecular ion (m/z = 58, since they are isomers), so it is the fragmentation pattern, not the molecular ion, that is needed to distinguish them
(c) Answer: CHO+ at m/z = 29; its presence indicates propanal, since propanone does not give this fragment.
(d) B1 infrared spectroscopy or 1H NMR spectroscopy named
(d) B1 IR: propanal shows an additional C-H (aldehyde) stretch near 2720-2850 cm-1, absent in propanone; OR NMR: propanal shows a characteristic aldehyde CHO proton signal at approx. 9.7-10 ppm, absent in propanone
(d) Answer: 1H NMR: propanal shows an aldehyde CHO proton at approx. 9.7-10 ppm, which propanone lacks (or IR: additional aldehyde C-H stretch near 2720-2850 cm-1).

Question 7

(a) B1 GC gives an automated, quantitative measurement (peak area/retention time), allowing concentration to be determined directly and precisely, unlike TLC which is mainly qualitative oe
(a) Answer: GC gives precise, quantitative peak area/retention time data, allowing concentration to be measured directly (TLC is mainly qualitative).
(b) B1 retention time is the time taken for a component to travel through (elute from) the GC column, from injection to detection
(b) B1 under fixed/identical conditions (same column, temperature, flow rate), each compound has a characteristic, reproducible retention time, so it can be compared with that of a known standard to help identify it (though this alone is not fully conclusive, e.g. co-elution is possible)
(b) Answer: Retention time is the elution time of a component; under fixed conditions it is characteristic of a compound, so can be compared with standards to aid identification.
(c) M1 rearrange: c = (780 - 15)/42.5
(c) dM1 = 765/42.5
(c) A1 18.0 mg per 100 cm3 (awrt 18.0)
(c) Answer: 18.0 mg per 100 cm3
(d) B1 below the legal limit (18.0 mg per 100 cm3 < 80 mg per 100 cm3)
(d) M1/A1 percentage = (18.0/80) x 100 = 22.5%
(d) Answer: Below the limit; 22.5% of the legal limit.
(e) B1 a known, fixed amount of internal standard is added to every sample and calibration standard, and the ratio of analyte peak area to internal standard peak area is used, rather than the absolute peak area
(e) B1 this corrects for/compensates for variations in injection volume or detector response between runs, improving the accuracy/reproducibility of the result
(e) Answer: Comparing analyte peak area to a fixed internal standard peak area corrects for run-to-run variation (e.g. injection volume), improving reliability.

Question 8

(a) B1 ester
(a) B1 the absorption at 1740 cm-1 corresponds to a C=O stretch, and the absence of a broad O-H absorption (2500-3300 cm-1) rules out a carboxylic acid, consistent with an ester
(a) Answer: Ester; C=O at 1740 cm-1 with no O-H absorption rules out a carboxylic acid.
(b) B1 three proton environments (singlet, quartet, triplet)
(b) B1 total H = 3 + 2 + 3 = 8, consistent with molecular formula C4H8O2
(b) Answer: 3 environments; total 8H, matching C4H8O2.
(c) B1 group lost: OCH2CH3 (ethoxy, OC2H5, mass 45)
(c) B1 ion formed: CH3CO+ (acylium ion, mass 43)
(c) Answer: OC2H5 (mass 45) lost; CH3CO+ (mass 43) formed.
(d) B1 structure: CH3COOCH2CH3 (ethyl ethanoate)
(d) B1 singlet at 2.0 ppm (3H) = CH3 bonded to C=O, with no adjacent protons, so no splitting
(d) B1 quartet at 4.1 ppm (2H) = OCH2, split by the adjacent CH3 (3H, n+1 = 4)
(d) B1 triplet at 1.3 ppm (3H) = CH3 of the ethyl group, split by the adjacent OCH2 (2H, n+1 = 3)
(d) Answer: Y is ethyl ethanoate, CH3COOCH2CH3.

Question 9

(a) B1 bromine occurs naturally as two isotopes, 79Br and 81Br, in an approximately 1:1 (50:50) natural abundance
(a) B1 the M peak (containing 79Br) and the M+2 peak (containing 81Br) therefore appear with roughly equal intensity
(a) Answer: Bromine's two isotopes (79Br, 81Br) occur in roughly equal abundance, giving M and M+2 peaks of similar height.
(b) B1 ratio = 9:6:1
(b) M1 with two chlorine atoms and a 3:1 (35Cl:37Cl) natural ratio, the probability of both being 35Cl is (3/4)² = 9/16, of one 35Cl and one 37Cl is 2 x (3/4)(1/4) = 6/16, and of both being 37Cl is (1/4)² = 1/16
(b) A1 giving the ratio 9:6:1 for M(35,35) : M+2(35,37) : M+4(37,37)
(b) Answer: M : M+2 : M+4 = 9 : 6 : 1
(c) M1 C3H8O: 3(12.0000) + 8(1.0078) + 15.9949
(c) A1 = 60.057 (awrt 60.057)
(c) M1 C2H4O2: 2(12.0000) + 4(1.0078) + 2(15.9949)
(c) A1 = 60.021 (awrt 60.021)
(c) B1 both compounds have the same nominal mass (60), so low-resolution MS cannot distinguish them; high-resolution MS measures mass to several decimal places, revealing the small difference (60.057 vs 60.021) and so identifies the correct molecular formula/compound
(c) Answer: C3H8O = 60.057; C2H4O2 = 60.021; high resolution MS distinguishes them by this small mass difference, which low resolution MS cannot detect.

Question 10

(a) B1 OCH3: singlet (3H)
(a) B1 CH2: quartet (2H)
(a) B1 CH3 (of ethyl group): triplet (3H)
(a) B1 correct reasoning for at least one pattern using the n+1 rule, e.g. CH2 is adjacent to the 3H of the ethyl CH3, so n=3 gives a quartet
(a) Answer: OCH3 singlet (3H); CH2 quartet (2H); CH3 triplet (3H).
(b) B1 in ethyl ethanoate the singlet (3H) is at a lower shift (approx. 2.0 ppm, CH3 next to C=O), whereas in methyl propanoate the singlet (3H) is at a higher shift (approx. 3.7 ppm, OCH3 next to O)
(b) B1 in ethyl ethanoate the CH2 quartet is next to oxygen (approx. 4.1 ppm, OCH2), whereas in methyl propanoate the CH2 quartet is next to the carbonyl carbon (approx. 2.3 ppm, COCH2), a lower shift
(b) B1 although the splitting patterns (singlet:quartet:triplet) are the same for both isomers, the precise chemical shift of each peak differs, allowing the two isomers to be distinguished
(b) Answer: The splitting patterns match, but the chemical shifts of the singlet and quartet differ between the two isomers, distinguishing them.
(c) B1 both methyl propanoate and ethyl ethanoate have four non-equivalent carbon atoms, so both give four peaks in their 13C NMR spectra; the peak count alone cannot distinguish them
(c) B1 comparing the precise 13C chemical shift values, or using 1H NMR data (e.g. as in part (b)), would allow the isomers to be distinguished
(c) Answer: Both isomers give four 13C peaks, so peak count is not enough; comparing exact shift values or using 1H NMR data would distinguish them.

Question 11

Level 3 (5-6): A full, logically structured answer that correctly identifies Z as 2-methylpropanal, (CH3)2CHCHO, using evidence from all three techniques (MS, IR and NMR) to justify both the functional group and the carbon skeleton, with clear reasoning linking specific data (m/z values, wavenumber, chemical shifts, integration, splitting) to structural features.
Level 2 (3-4): Correctly identifies most structural features (e.g. the aldehyde group and/or the branched carbon skeleton) using evidence from at least two techniques, with generally sound reasoning, but the final structure may be incomplete, not fully justified, or contain minor errors or omissions in linking evidence to structure.
Level 1 (1-2): Some relevant points made, e.g. correctly identifies the functional group present or interprets isolated pieces of data, but with little or no attempt to combine evidence from different techniques; the final structure is incorrect or absent.
Level 0 (0): No relevant content.
Indicative content:
- IR: absorption at 1715 cm-1 indicates a C=O bond; absence of a broad O-H absorption (2500-3550 cm-1) rules out a carboxylic acid, so an aldehyde or ketone is indicated.
- MS: the fragment at m/z = 29 corresponds to CHO+ (formyl/acylium cation), which is characteristic of an aldehyde, confirming the C=O is terminal (an aldehyde) rather than a ketone.
- MS: the molecular ion at m/z = 72 confirms Mr = 72, consistent with molecular formula C4H8O.
- MS: the fragment at m/z = 57 corresponds to loss of CH3 (mass 15) from the molecular ion (72 - 15 = 57), i.e. loss of a methyl group.
- NMR: the doublet at 1.15 ppm, integration 6H, indicates two equivalent CH3 groups, each adjacent to only one proton (n+1 rule, n=1 gives a doublet); the 6H integration confirms two equivalent methyl groups, consistent with an isopropyl-type (CH3)2CH- unit.
- NMR: the multiplet at 2.55 ppm (1H) is the CH proton, coupled to both methyl groups and to the aldehyde proton, consistent with a CH bonded to two CH3 groups and to a CHO group.
- NMR: the doublet at 9.60 ppm (1H) is the aldehyde CHO proton (characteristic shift 9.5-10 ppm for aldehydes), split into a doublet by the single adjacent CH proton.
- Combining all the evidence gives the structure 2-methylpropanal (isobutyraldehyde), (CH3)2CHCHO, with a correct displayed or skeletal formula consistent with every data point given.

Question 12

(a) M1/A1 5400/3600 = 1.50 (cao)
(a) Answer: 1.50
(b) M1 rearrange: mass X = (area ratio / k) x mass IS
(b) M1 substitute: (1.50/1.20) x 50.0
(b) A1 = 62.5 ng (awrt 62.5)
(b) Answer: 62.5 ng
(c) M1 62.5/2.00
(c) A1 = 31.25 ng per cm3 (awrt 31.3)
(c) Answer: 31.25 ng per cm3 (awrt 31.3)
(d) B1 the concentration (31.25/31.3 ng per cm3) exceeds the 30 ng per cm3 threshold, so the sample would be reported/flagged as positive
(d) B1 a confirmatory test using different chromatographic/MS conditions reduces the chance of a false positive caused by a co-eluting or interfering compound with a similar retention time or fragmentation pattern
(d) B1 this increases confidence/certainty in the result, given the serious consequences of an adverse finding for the athlete
(d) Answer: Positive (31.25 ng/cm3 > 30 ng/cm3 threshold); a confirmatory test with different conditions guards against false positives from interfering compounds, which matters given the serious consequences for the athlete.
(e) B1 the internal standard corrects for variations between runs (e.g. differences in injection volume, extraction efficiency or detector response), because both X and IS are affected equally/proportionally
(e) B1 using the ratio of peak areas (rather than the absolute area of X alone) cancels out these systematic errors, giving a more reliable/reproducible measurement of concentration
(e) Answer: The internal standard cancels out run-to-run variation, since it affects both X and IS proportionally, making the peak-area ratio more reliable than an absolute area.

Analytical Techniques - Worksheets, Questions and Revision