A-LevelChemistryAnalytical techniques9701/41 · Oct/Nov 2022 · Q912 Marks

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. [Figure 9.1 Chromatogram showing peaks A, B, C at different retention times]

(a)(i)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 [1][1]
(a)(ii)Explain the meaning of retention time. [1][1]
(a)(iii)Calculate the percentage of B in the mixture. Show your working. percentage of B = ...................................... % [2][2]
(b)Complete [Table 9.1] 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 with structural formulae and empty 'number of peaks' column] [2][2]
(c)State the number of peaks that would be seen in the proton (1H) NMR spectrum of methyl butanoate, CH3CH2CH2CO2CH3. Name all the splitting patterns seen in this spectrum. number of peaks splitting patterns [2][2]
(d)(i)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]. [Figure 9.2 showing 1H NMR spectrum D] [Figure 9.3 showing 1H NMR spectrum E] [Table 9.2 showing typical chemical shifts for various proton environments] Deduce the structures of the two esters D and E and draw their displayed formulae in the boxes below. [Empty boxes for D and E structures] [2][2]
(d)(ii)The spectrum of D includes a quartet at δ 4.1. Identify the protons responsible for this quartet on your structure in (i) by labelling these protons with the letter F. Explain why this peak is split into a quartet. [1][1]
(d)(iii)The spectrum of E has a doublet at δ 1.1. Identify the protons responsible for this doublet on your structure in (i) by labelling these protons with the letter G. Explain why this peak has a chemical shift of 1.1. [1][1]

This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2022). The mark or mark type is shown in brackets.

[1] stationary: non-volatile (non-polar) liquid mobile: nitrogen / argon / any inert gas
[1] the time that the substance stays in the column OR time between injection and detection OWTTE
[2] areas are 30, 100, 25 OR 30, 150, 25 (½ × b × h) OR 100 / 155 [1] % of B = 100 × (100 / 155) OR % of B = 100 × 150 / 205 = = 64.5 = 73.2 [1] ecf M1 areas
[2] CH3CH2CH2CO2CH3: 5 CH3CH2CO2CH2CH3: 5 CH3CO2CH2CH2CH3: 5 (CH3)2CHCO2CH3: 4 CH3CO2CH(CH3)2: 4
[2] 4 [1] singlet, (two) triplet(s), multiplet (any order) [1]
[2] D = CH3CH2CO2CH2CH3 [1] E = (CH3)2CHCO2CH3 [1]
[1] O-CH2 labelled F AND three protons on neighbouring carbon / adjacent CH3
[1] both CH3 in isopropyl group labelled G AND alkane / alkyl (protons)