OCR A Specimen Paper 2 Q21 (with explained solutions) |

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OCR A Specimen Paper 2 Q21

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21* A chemist isolates compound L, with empirical formula C3H6O, and sends a sample for analysis. 32 The analytical laboratory sends back the following spectra. Mass spectrum Molecular ion peak at m/z = 116.0. 1H NMR spectra The numbers next to each signal represent the number of 1H responsible for that signal. Two 1H NMR spectra were obtained: one without D2O and one with D2O added. 1H NMR spectrum with no D2O: Expansion of multiplet centred at2.7 ppm Specimen 1H NMR spectrum with D2O added:OCR 2014 H432/02<br />
13C NMR spectrum: 33 Use the information provided to suggest a structure for compound L. Give your reasoning. Specimen . . . . . . . . . .OCR 2014 H432/02 END OF QUESTION PAPER [6]<br />

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H432/02 Question 21* Mark Scheme June 20XX Guidance Indicative scientific points may be included: Structure Marks Answer L = Please refer to the marking instructions on page 4 of this mark scheme for guidance on how to mark this question. Level 3 (56 marks) Structure correct AND Analysed all 1H NMR signals with at least two supporting statements made. The analysis is clear and logically structured. The supporting statements are relevant to the correct structure drawn. Level 2 (34 marks) Structure has correct molecular formula AND C=O AND OH but in incorrect positions AND Analysed at least three 1H NMR signals with one or two supporting statements made The analysis is presented with some structure. The supporting statements are in the most-part relevant to the structure drawn. Level 1 (12 marks) Structure has correct molecular formula AND C=O OR OH but in incorrect positions AND Analysed at least two 1H NMR signals with no or one supporting statements made The analysis is basic and communicated in an unstructured way. The relationship of the supporting evidence to the structure may not be clear. Specimen 1H NMR spectrum= 3.8 ppm, triplet, 2H= 3.7 ppm, singlet, 1H= 3.1 ppm, triplet, 2H= 2.7 ppm, septet, 1H= 1.0 ppm, doublet, 6H CH2CH2O OH CH2CH2C=O (CH3)2CHC=O (CH3)2CH Supporting statements= 3.7 ppm lost after D2O, indicating OH= 213 ppm in 13C NMR but no= 910 ppm in 1H NMR so ketone, not aldehyde Mr(C3H6O) = 58 116/58 = 2C6H12O2 21 P M T<br />
H432/02 Question 0 marks No response or no response worthy of credit. Answer Mark Scheme June 20XX Marks Guidance Total Specimen 22 P M T<br />

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