AQA AS Specimen Paper 2 Q1 (with explained solutions) |

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AQA AS Specimen Paper 2 Q1

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Section A Answer all questions in this section. Compound J, known as leaf alcohol, has the structural formula CH3CH2CH=CHCH2CH2OH and is produced in small quantities by many green plants. The E isomer of J is responsible for the smell of freshly cut grass. Give the structure of the E isomer of J. [1 mark] Give the skeletal formula of the organic product formed when J is dehydrated using concentrated sulfuric acid. [1 mark] . 1 0 1 . 2 0 1 Barcode Typesetter code<br />
. 3 Another structural isomer of J is shown below. CH3CH2 CH3 C C CH2OH 0 1 Explain how the Cahn-Ingold-Prelog (CIP) priority rules can be used to deduce the full IUPAC name of this compound. [6 marks] Question 1 continues on the next page Barcode Typesetter code Turn over <br />
0 1 . 4 The effect of gentle heat on maleic acid is shown below. A student predicted that the yield of this reaction would be greater than 80%. In an experiment,10.0 g of maleic acid were heated and 6.53 g of organic product were obtained. Is the student correct? Justify your answer with a calculation using these data. [2 marks] Barcode Typesetter code<br />
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Question 01.1 CH3CH2 Marking guidance 01.2 CH2CH2OH MARK SCHEMEAS ChemistrySpecimen paper 2 Comments Mark AO AO1a AO2c 2 of 14<br />
Stage 1: consider the groups joined to right hand carbon of the C=C bond Consider the atomic number of the atoms attached C has a higher atomic number than H, so CH2OH takes priority Stage 2: consider the groups joined to LH carbon of the C=C bond Both groups contain C atoms, so consider atoms one bond further away C, (H and H) from ethyl group has higher atomic number than H, (H and H) from methyl group, so ethyl takes priority Stage 3: conclusion The highest priority groups, ethyl and CH2OH are on same side of the C=C bond so the isomer is Z The rest of the IUPAC name is 3-methylpent-2-en-1-ol 1 1 1 1 1 1 01.3 AO1a AO2a AO2a AO2a AO2a AO2a MARK SCHEMEAS ChemistrySpecimen paper 2 Extended response Maximum of 5 marks for answers which do not show a sustained line of reasoning which is coherent, relevant, substantiated and logically structured. M1 can be scored in stage 1 or stage 2 Allow M5 for correct ECF conclusion using either or both wrong priorities deduced in stages 1 and 2 3 of 14<br />
MARK SCHEMEAS ChemistrySpecimen paper 2 01.4 Moles of maleic acid = 10.0/116.0 = 8.62102 AND mass of organic product expected = (8.62102 )98.0 = 8.45 g Or moles of organic product formed = 6.53 / 98.0 = 6.66102 % yield = 1006.53/8.45 OR = 100(6.66102) / (8.62102) AND statement that the student was NOT correct = 77.294 = 77.3% 1 1 AO3 1a AO3 1a 4 of 14<br />

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