The problems are on the spectroscopy worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Mass spectrometry gives the relative molecular mass from the molecular ion and the structure from fragments, infrared spectroscopy identifies functional groups from bond absorptions, and proton and carbon-13 NMR show how many hydrogen and carbon environments exist and how they are joined. These worked answers correct the 20 problems of the Chemistry 11-12 worksheet on spectroscopy, from molecular ion of ethanol to propanol isomers in forensics: each solution shows the formula, the substitution and the final answer with its unit.
Each answer links the data to the conclusion, with calculated values in bold. If your structure differs, check each piece of evidence against the data again, because one missed neighbour count often changes the answer.
Level: Chemistry grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB S3.2
Problem 1 Molecular ion of ethanol
1. M⁺ equals the relative molecular mass of the compound.2. 2 × 12 + 6 × 1 + 16 = 24 + 6 + 16 = 46.
3. CH₂OH⁺: 12 + 2 + 16 + 1 = 31, so it explains the peak, formed by losing CH₃ from the molecular ion.
Problem 2 Matching IR bands
1. 3350 cm⁻¹ = O–H (broad, alcohol); 1715 cm⁻¹ = C=O; 2240 cm⁻¹ = C≡N; 1650 cm⁻¹ = C=C.2. None of them: the fingerprint region is below 1500 cm⁻¹.
Problem 3 Counting hydrogen environments
1. CH₃CH₂CH₃: 2; CH₃OCH₃: 1; CH₃CH₂OH: 3; benzene: 1.2. The two methyl groups are bonded to the same oxygen, so all six hydrogens are in the same environment.
Problem 4 Predicting the splitting
1. CH₃ (3H) has one neighbour, so a doublet of 3H; CH (1H) has three neighbours, so a quartet of 1H.2. A singlet of 4H: all four hydrogens are equivalent, and equivalent hydrogens do not split each other.
Problem 5 Wavelength to wavenumber
1. 5.88 μm = 5.88 × 10⁻⁴ cm, so 1 ÷ 5.88 × 10⁻⁴ = 1700 cm⁻¹. 3.00 μm = 3.00 × 10⁻⁴ cm, so 1 ÷ 3.00 × 10⁻⁴ = 3330 cm⁻¹.2. 1700 cm⁻¹ is C=O; 3330 cm⁻¹ is O–H.
Problem 6 Counting carbons from M+1
1. Ratio = 2.4 ÷ 55.0 = 4.36 %; 4.36 ÷ 1.1 = 3.97, so 4 carbon atoms.2. 88 − 4 × 12 − 2 × 16 = 88 − 48 − 32 = 8, so the formula is C₄H₈O₂.
Problem 7 Two ketone isomers
1. CH₃CH₂CO⁺: 36 + 5 + 16 = 57; C₂H₅⁺: 24 + 5 = 29.2. It is pentan-3-one, CH₃CH₂COCH₂CH₃, which breaks on either side of C=O to give C₂H₅CO⁺ and C₂H₅⁺.
3. Pentan-2-one would give 43 (CH₃CO⁺) and 71 (loss of CH₃).
Problem 8 Alcohol or ketone
1. Propanone: strong C=O at 1715 cm⁻¹ and no O–H.2. Prop-2-en-1-ol would show a broad O–H absorption at 3230–3550 cm⁻¹ and a C=C absorption near 1650 cm⁻¹.
Problem 9 13C environments
1. Pentane: CH₃, CH₂ (next to CH₃), middle CH₂: 3.2. Propanone: CH₃ and C=O: 2.
3. 2-methylpropane: three equivalent CH₃ and one CH: 2.
4. Ethyl ethanoate: CH₃CO, C=O, OCH₂ and CH₃: 4.
5. 1,4-dimethylbenzene: CH₃, ring C–CH₃ and ring CH: 3.
Problem 10 Propanoic acid data
1. A carboxylic acid: very broad O–H and C=O at 1710 cm⁻¹.2. δ 11.5 singlet 1H = acid O–H; δ 2.38 quartet 2H = CH₂ next to C=O with 3 neighbouring H from CH₃; δ 1.16 triplet 3H = CH₃ with 2 neighbours.
3. CH₃CH₂COOH: 36 + 6 + 32 = 74 checks. Propanoic acid.
Problem 11 Reading integration traces
1. 24 : 16 = 3 : 2.2. Total 10 H over 5 parts = 2 H per part: 6 H and 4 H.
3. The triplet (6H, two equivalent CH₃) next to CH₂ and the quartet (4H, two equivalent CH₂ at δ 3.5 next to oxygen) with no O–H absorption give ethoxyethane, CH₃CH₂OCH₂CH₃.
Problem 12 Why TMS is the reference
1. TMS is Si(CH₃)₄.2. All 12 hydrogens are equivalent, so it gives one sharp signal at δ = 0, away from most signals; it is inert and volatile, so it is easily removed.
3. CDCl₃ has deuterium instead of hydrogen, so the solvent gives no proton signal that would hide the sample’s signals.
Problem 13 Halogen isotope patterns
1. ³⁵Cl compound: C₃H₇³⁵Cl = 36 + 7 + 35 = 78; ³⁷Cl compound: 80. Heights 78 : 80 = 3 : 1.2. C₄H₉⁷⁹Br = 48 + 9 + 79 = 136; C₄H₉⁸¹Br = 138. Heights 1 : 1.
Problem 14 An alcohol from its IR
1. The broad absorption is the O–H bond of an alcohol, broadened by hydrogen bonding.2. Ethoxyethane has no O–H, so it could not give an absorption near 3340 cm⁻¹.
3. The 1060 cm⁻¹ band lies in the fingerprint region, where many C–O and C–C absorptions overlap; it matches a compound only against a reference spectrum, and cannot name the bond.
Problem 15 Fragments of a small aldehyde
1. The molecular ion is at m/z 44 (24 + 4 + 16 = 44).2. 15 = CH₃⁺; 29 = CHO⁺ (12 + 1 + 16); 43 = [M − H]⁺, that is CH₃CO⁺ (24 + 3 + 16 = 43).
3. IR at 1730 cm⁻¹ shows C=O, so the compound is ethanal, CH₃CHO.
Problem 16 A full ester analysis
1. IR 1738 cm⁻¹ and no O–H: an ester (C=O at 174 ppm supports this).2. δ 4.12 quartet 2H = O–CH₂ next to a CH₃; δ 1.25 triplet 3H = the CH₃ of that ethyl group; δ 2.31 quartet 2H = CH₂ next to C=O and CH₃; δ 1.12 triplet 3H = the CH₃ of the propanoyl group.
3. Ethyl propanoate, CH₃CH₂COOCH₂CH₃ (60 + 10 + 32 = 102). Five different carbons (two CH₃, two CH₂, one C=O) give 5 signals, as observed; fragments 57 (C₂H₅CO⁺) and 29 (C₂H₅⁺) agree.
Problem 17 A nitrogen compound
1. M = 55 with C, H, N: C₃H₅N = 36 + 5 + 14 = 55, so C₃H₅N.2. A sharp absorption at 2250 cm⁻¹ is the C≡N bond.
3. Triplet 3H + quartet 2H means an ethyl group, so the structure is CH₃CH₂C≡N, propanenitrile; the 13C signal near 120 ppm is the nitrile carbon, and 3 carbons give 3 signals.
Problem 18 A bromoalkane with a single carbon branch
1. 5.5 ÷ 1.1 = 5 carbon atoms.2. Equal M and M+2 (150 and 152) show one bromine. 150 − 79 = 71; 71 − 60 = 11, so the formula is C₅H₁₁Br.
3. 1-bromo-2,2-dimethylpropane, (CH₃)₃CCH₂Br. Nine equivalent H (δ 1.05) sit on three CH₃ groups attached to a quaternary carbon, and the two H of CH₂Br (δ 3.3) have no hydrogens on the adjacent carbons, so both signals are singlets. Three 13C signals: CH₃, central C and CH₂Br.
Problem 19 Telling aldehyde isomers apart
1. Butanal: 4 (CHO, CH₂, CH₂, CH₃); butanone: 4 (CH₃CO, C=O, CH₂, CH₃); 2-methylpropanal: 3 (two equivalent CH₃, CH, CHO).2. 2-methylpropanal, (CH₃)₂CHCHO: 6H doublet (2 × CH₃, one neighbour), 1H multiplet (CH), 1H doublet (CHO).
3. The CHO hydrogen has one neighbouring hydrogen (on the CH carbon), so the n + 1 rule gives a doublet.
Problem 20 Propanol isomers in forensics
1. CH₂OH⁺: 12 + 2 + 16 + 1 = 31; CH₃CHOH⁺: C₂H₅O⁺ = 24 + 5 + 16 = 45.2. P (base peak 31) is propan-1-ol; Q (base peak 45) is propan-2-ol.
3. Both contain O–H, C–H and C–O, so absorptions above 1500 cm⁻¹ are almost the same. 13C NMR gives 3 signals for propan-1-ol and 2 signals for propan-2-ol.
