OCR A Jun 2014 Paper 2 Q8 (with explained solutions) |

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OCR A Jun 2014 Paper 2 Q8

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(a) Compound B, CxHyO, can be oxidised to form a ketone C. 18 0.035 mol of B has a mass of 2.59 g. Compound B reacts with compound D, C3H6O2, in the presence of an acid catalyst to form two compounds E and F.Calculate the molar mass of compound B.Give the structures of compounds B, C, D, E and F.[6]OCR 2014<br />
19 BLANK PAGE Question 8 continues on page 20 PLEASE DO NOT WRITE ON THIS PAGEOCR 2014 Turn over<br />
20 (b) Compound G is a branched-chain organic compound that does not have E and Z isomers. Elemental analysis of compound G gave the following percentage composition by mass: C, 55.8%; H, 7.0%; O, 37.2%. The mass spectrum and infrared spectrum of compound G are shown below. 20 30 40 50 60 m / z 70 80 90 100 Mass spectrum 100 relative intensity 80 60 40 20 10 Infrared spectrum 100 transmittance (%) 50 0 4000 3000OCR 2014 2000 wavenumber / cm1 1500 1000 500<br />
Calculate the empirical and molecular formulae for compound G. 21Write the formulae for the particles responsible for peak X and peak Y in the mass spectrum.Draw the structure of compound G. Explain fully how you arrive at a structure for compound G using all the evidence provided.[7] [Total: 13]OCR 2014 END OF QUESTION PAPER<br />

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F322 Question 8 (a) Molar mass of B = 74 (cid:57) B-F clearly identified B/alcohol: Answer Mark Scheme Mark C/ketone: OH D/carboxylic acid: (cid:57) (cid:57) E and F: H2O/water (cid:57) (cid:57) CH3 (cid:57) June 2014 Guidance ANNOTATE ANSWER WITH TICKS AND CROSSES ETC Check and annotate page 19 below this response Molar mass = 2.59 = 74 0.035 For structure of B, C, D or E/F ALLOW correct displayed OR correct structural formula OR correct skeletal formula OR mixture of the above as long as unambiguous. DO NOT ALLOW missing H atom(s) in a displayed formula for one structure but ALLOW missing H atoms in subsequent structures. IGNORE names of organic compounds E and F can be identified either way round ALLOW H2O or displayed formula for mark For E and FALLOW the two optical isomers C2H5 CH3 H3C C2H5 22 C2H5 C2H5<br />
F322 Question 8 (b) Answer Mark Scheme June 2014 Molecular formula for G: 8.55 Mole ratio C : H : O = 0.12 0.7 0.1 2 marks 2.37 0.16 2 marks + (cid:57) OR 4.65 : 7.0 : 2.33/2.325 OR 2 : 3 : 1 OR C2H3O (cid:57) Molecular formula of G C4H6O2 (cid:57) Mass spectrum for G: Peak X or peak 41 indicates C3H5 Peak Y or peak 45 indicates COOH+ (cid:57) Infrared for G: 1 mark Peak at 16401750 cm-1 indicates presence of C=O AND Peak at 25003300 cm-1 (indicates the presence of) OH group linked carboxylic acid/COOH QWC (cid:57) Mark Guidance ANNOTATE ANSWER WITH TICKS AND CROSSES ETC ALLOW mass of C = 0.558 x 86 or 48 AND mass of H = 0.07 x 86 or 6 AND mass of O = 0.372 x 86 = 32 + charge required for each response ALLOW one mark if both formulae are correct but with no charge/incorrect charge ALLOW any possible fragments that contain C, H and/or O that have the correct mass. E.g. Peak X indicates C2OH+, Peak Y indicates C2H5O+ Unfeasible fragments are not allowed e.g. C3H9 atoms) LOOK ON THE SPECTRUM for labelled absorbance which can be given credit Candidates must link absorbance to bond in order to gain the mark ALLOW 1700 cm-1 For 25003300 cm-1, ALLOW 2900 cm-1 or any stated wavenumber with range 25003300 cm-1 ALLOW wavenumber range up to 24003500 cm-1 + (too many H 23<br />
F322 Question Mark Scheme June 2014 Answer Structure of G: 2 marks Correct structure: H CH3 1 mark for one of the following structures of C4H6O2: H (cid:57)(cid:57) Mark Guidance ALLOW structural, skeletal or displayed formula. DO NOT ALLOW ECF from incorrect molecular formula (cid:57) OR OR Total 13 24<br />

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