OCR A Jun 2013 Paper 4 Q4 (with explained solutions) |

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OCR A Jun 2013 Paper 4 Q4

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4 Compound E, shown below, is an alcohol that is found in oak wood. It is formed by the breakdown of cellulose. 12 OH OH compound E Compound E can be converted into compound G as shown in the flowchart below. OH OH excess K2Cr2O7/H2SO4 reflux compound E reaction 1 compound F reaction 2 reagentOH OH compound G (a) (i) Complete the flowchart to show the structure of the organic compound F and the reagent [1] needed for reaction 2.OCR 2013<br />
(ii) What would you observe during reaction 1? 13[1] (iii) In reaction 1, compound E was heated under reflux with excess K2Cr2O7/H2SO4. Suggest why these conditions were used, rather than the reaction mixture being distilled during the process.[1] (iv) Name the type of reaction taking place in reaction 2.[1] (b) Describe a chemical test that you could use to detect the presence of a carbonyl group in an organic compound. ReagentObservation[2] (c) Compound E is a single stereoisomer. (i) Draw the skeletal formula of one other stereoisomer of compound E and state the type of stereoisomerism. type of stereoisomerism[2]OCR 2013 Turn over<br />
(ii) 4.56 g of compound E was converted into compound G using the method shown in the flowchart on page 12. 14 3.15 g of compound G was formed. Calculate the percentage yield of compound G. Give your answer to three significant figures. The Mr of compound E is 160.0. percentage yield of compound G =% [3] (iii) Compound G is heated for several hours under reflux, in the presence of a concentrated sulfuric acid catalyst. An ester and a small inorganic molecule are formed. Complete the equation below to show the two products formed by this reaction. concentrated H2SO4 catalyst OH OH heat under reflux +[2] [Total: 13]OCR 2013<br />

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F324 Question (i) (a) Answer F = AND reagent NaBH4 (cid:57) NB One mark for BOTH Mark Scheme Marks Guidance ALLOW correct structural OR displayed OR skeletal formulae OR combination of above as long as unambiguous June 2013 Wedge out of the paper is required i.e.( DO NOT ALLOW dashed wedge on methyl group in this orientation or or or or or ( ALLOW O OH ALLOW redox OR reduction ALLOW Bradys (reagent) ALLOW orange/red/yellow for colour of the 2,4-DNP(H) precipitate ALLOW solid/crystals in place of precipitate IGNORE any reference to recrystallising/melting points (ii) Colour changes from orange to green / blue / green blue (cid:57) to ensure carboxylic acid is formed OR prevents formation of aldehyde OR distillation only makes the aldehyde (cid:57) (nucleophilic) addition (cid:57) (iii) (iv) (b) 2,4-DNP(H) orange precipitate (cid:57) (cid:57)(cid:3)<br />
F324 Question (c) (i) One of: HO HO OR OR Mark Scheme Marks Answer June 2013 Guidance or or or For bold wedge ALLOW For dashed wedge ALLOW DO NOT ALLOW any other representation of the structure, i.e. anything not skeletal ALLOW open wedges ALLOW isomers shown in any alternative correct orientation or or (cid:57) for one mark optical (isomerism) (cid:57) If answer = 63.5 award 3 marks moles of E used = 4.56/160(.0) / 0.0285 (mol) (cid:57) moles of G formed = 3.15/174(.0) / 0.0181 (mol) (cid:57) yield = 0.0181/0.0285100% / 63.5% (cid:57) (ii) 10 0.0285 mol is exact calculator value 0.0181 mol is to 3sf (calculator value 0.0181034) IGNORE trailing numbers in this answer ALL ANSWERS MUST be to a minimum of 3sf, the final answer must be to 3 sf (calculator value gives 63.520871%) (rounding of moles of G gives 63.508772%) ALLOW ecf from incorrect Mr or moles unless the yield is >100%<br />
F324 Question (iii) Mark Scheme June 2013 Answer Marks Guidance ALLOW abbreviation of alkyl chain Other product = H2O for first mark (cid:57) for second mark (cid:57) Wedge out of the paper is required i.e.( DO NOT ALLOW dashed wedge on methyl group in this orientation or or or or or ( ALLOW Be careful with orientation of lactone: ALLOW Total 13 11<br />

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