AQA Jun 2017 Paper 2 Q10 (with explained solutions) |

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AQA Jun 2017 Paper 2 Q10

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201 0 1 0 . 1 This question is about six isomers of C6H10O2 Give the full IUPAC name of isomer P. [1 mark] A sample of P was mixed with an excess of oxygen and the mixture ignited. After cooling to the original temperature, the total volume of gas remaining was 335 cm3 When this gas mixture was passed through aqueous sodium hydroxide, the carbon dioxide reacted and the volume of gas decreased to 155 cm3 Both gas volumes were measured at 25 C and 105 kPa Write an equation for the combustion of P in an excess of oxygen and calculate the mass, in mg, of P used. The gas constant R = 8.31 J K1 mol1 [5 marks] 1 0 . 2 *20* Mass of P used mg IB/M/Jun17/7405/2<br />
21Figure 4 Isomer Q (C6H10O2) is a cyclic compound. The infrared spectrum of Q is shown in Figure 4 and the 13C NMR spectrum of Q is shown in Figure 5. Use these spectra and Tables A and C in the Data Booklet to deduce the structure of Q. In your answer, state one piece of evidence you have used from each spectrum. Figure 5 [3 marks] 1 0 . 3 *21* Structure of Q. Evidence from Figure 4 Evidence from Figure 5 Turn overIB/M/Jun17/7405/2<br />
Although the 13C spectra of R and S both show the same number of peaks, the spectra can be used to distinguish between the isomers. Justify this statement using Table C from the Data Booklet. Give the number of peaks for each isomer. [3 marks] 22Isomers R and S are shown. 1 0 . 4 Justification Number of peaks *22* IB/M/Jun17/7405/2<br />
23Although the 1H spectra of R and S both show the same number of peaks, the spectra can be used to distinguish between the isomers. Justify this statement using the splitting patterns of the peaks. Give the number of peaks for each isomer. [3 marks] Justification 1 0 . 5 Number of peaks Question 10 continues on the next page *23* Turn overIB/M/Jun17/7405/2<br />
241 0 . 6 The action of heat on 5-hydroxyhexanoic acid can lead to two different products. On gentle heating, 5-hydroxyhexanoic acid loses water to form a cyclic compound, T (C6H10O2). Under different conditions, 5-hydroxyhexanoic acid forms a polyester. Draw the structure of T. Draw the repeating unit of the polyester and name the type of polymerisation. [3 marks] Structure of T Repeating unit of polyester Type of polymerisation *24* IB/M/Jun17/7405/2<br />
251 0 . 7 Isomer U is shown. The polymer formed by U and the polymer formed by 5-hydroxyhexanoic acid in Question 10.6 both contain ester groups that can be hydrolysed. Draw the repeating unit of the polymer formed by U. Justify the statement that, although both polymer structures contain ester groups, the polymer formed by U is not biodegradable. [3 marks] Repeating unit of polymer formed by U. Justification Turn over for the next question 21 *25* Turn overIB/M/Jun17/7405/2<br />

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MARK SCHEMEA-LEVEL CHEMISTRY7405/2JUNE 2017 Answers Mark Additional Comments/Guidance Z-2-methylpent-2-en (-1-) oic acid Ignore missing hyphens or extra commas, spaces, hyphens C6H10O2 + 7 O26CO2 + 5H2O Volume of CO2 formed = 180 cm3 Mol carbon dioxide = pV/RT = 105000(18010-6) Mol P, C6H10O2 used = 7.63210-3 / 6 = 1.27210-3 Mass P used = 7.63210-3 = 1.27210-3114(.0) g 8.31298 = 145 mg M1 M2 M3 M4 M5 Allow multiple If incorrect volume:155 gives 125mg / 335 gives 270mg could score M1,M3,M4max 3 If incorrect volume from AE then penalise M2 and mark on (Final answer is 0.806 x their volume) If unit error in p, V or T lose M3 and M5 If incorrect rearrangement lose M3 and M5 If both errors seen then no further marks M3 divided by 6 If wrong no further marks Mark for answer (allow ans to 2 sf) Check chemical equation before awarding final mark 28 of 32 Question 10.1 10.2<br />
MARK SCHEMEA-LEVEL CHEMISTRY7405/2JUNE 2017 10.3 10.4 OR Mark independently M1 Apply the list principle Fig 4: IR OH (acid) peak (2500-3000cm-1) present M2 Ignore C=O signal at 1750 cm-1 Fig 5:13C NMR 4 peaks so 4 (non-equivalent) environments Or Peak at 160-185 (show C=O) in (esters or) acids Or Peak at 40-50 (show R-CO-CH) presence of carbonyl Both M2 & M3 can be awarded on the spectra Allow correct Fig 4 answers in Fig 5 and converse M3 R has 4 C next to C=O S has 2 C next to C=O in range= 20-50 R has two peaks and S only one peak in this range Or R has more peaks (allowed if no numbers given) OR S has a -C(H2)-C(H3) R does not S has one peak in range= 5-40 R does not / lowest peak for S is lower than lowest for R (Both have) three peaks M1 for structural point M2 for resulting peak in spectra M1 M2 M1 M2 M3 29 of 32<br />
MARK SCHEMEA-LEVEL CHEMISTRY7405/2JUNE 2017 R Both singlets S has triplet and a quartet OR R CH3/peak at 2.1-2.6 is a singlet 10.5 S CH3/peak at 0.7-1.2 is a triplet OR R CH2/peak at 2.1-2.6 is a singlet S CH2/peak at 2.1-2.6 is a quartet (Both have) two peaks M1 M2 M1 M2 M1 M2 M3 10.6 OR OR Must have trailing bonds Ignore brackets and n condensation Ignore esterification 30 of 32<br />
MARK SCHEMEA-LEVEL CHEMISTRY7405/2JUNE 2017 Strong / non-polar C-C bonds (in the chain) cannot be attacked by nucleophiles/acids/cannot be hydrolysed. 10.7 OR Only polar ester group Can be attacked by nucleophiles/acids/can be hydrolysed Total M1 M2 M3 M2 M3 21 Must have trailing bonds Ignore brackets and n M3 dependent on correct or close M2 Allow 1 mark for in (polar) ester link in side chain/not in main chain therefore polymer chain not broken 31 of 32<br />

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