The questions are on the spectroscopy test. Take it under real conditions before reading the mark scheme.
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. This mark scheme gives the model answers and the marks for the 8 questions of the Chemistry 11-12 test on spectroscopy, from reading a carbonyl band to counting carbons, with the equivalent answers to accept.
Model answers and marking points follow. Give credit for each correct step, allow error carried forward in a later part, and do not award marks for a correct answer with a contradicting reason.
Level: Chemistry grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB S3.2
Question 1 Reading a carbonyl band (2 marks)
a. C=O.
b. A carboxylic acid has a very broad O–H absorption from 2500 to 3300 cm⁻¹, which is absent.
Marking: 1 for C=O; 1 for absence of O–H.
Question 2 Halogen molecular ions (3 marks)
a. C₃H₇⁷⁹Br: 36 + 7 + 79 = 122; C₃H₇⁸¹Br: 36 + 7 + 81 = 124.
b. 1 : 1, because the two bromine isotopes are almost equally abundant.
Marking: 1 for each m/z; 1 for ratio with reason.
Question 3 Tertiary alcohol NMR (3 marks)
a. 2 signals, ratio 9 : 1 (three equivalent CH₃ and one OH).
b. The CH₃ hydrogens have no hydrogens on adjacent carbons, and the OH hydrogen does not split, so neither is split.
Marking: 1 for two signals; 1 for 9 : 1; 1 for reason.
Question 4 Hexane isomers in 13C NMR (3 marks)
a. Hexane: 3.
b. 2,2-dimethylbutane: 4 (three equivalent CH₃, quaternary C, CH₂, CH₃).
c. 2,3-dimethylbutane: 2 (four equivalent CH₃ and two equivalent CH).
Marking: 1 mark each.
Question 5 Identify a small ester (3 marks)
a. The compound has two CH₃ groups, one next to C=O and one next to oxygen, and neither has hydrogens on adjacent atoms, so each is a singlet; each has 3 H, so the heights are equal.
b. Methyl ethanoate, CH₃COOCH₃.
Marking: 1 for no adjacent hydrogens; 1 for 3H : 3H; 1 for methyl ethanoate.
Question 6 Wavenumber and bond strength (2 marks)
a. 1 ÷ 1715 = 5.83 × 10⁻⁴ cm = 5.83 μm.
b. The C=O bond is stronger than the C=C bond, so it vibrates at higher frequency.
Marking: 1 for 5.83 μm; 1 for stronger bond.
Question 7 Deuterium oxide shake (2 marks)
The O–H signal disappears. The O–H hydrogen exchanges with deuterium to give O–D, and deuterium does not absorb in proton NMR.
Marking: 1 for disappears; 1 for exchange with deuterium.
Question 8 Counting carbons (2 marks)
3.08 ÷ 40.0 = 7.7 %; 7.7 ÷ 1.1 = 7.0, so 7 carbon atoms.
Marking: 1 for 7.7 %; 1 for 7.
