OCR A AS Specimen Paper 1 Q22 (with explained solutions) |

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OCR A AS Specimen Paper 1 Q22

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22 The organic compounds labelled A to E below are all produced by living organisms. 18 (a) State the systematic name of compound A. SPECIMEN (b) Compound D reacts readily with hydrogen chloride in an addition reaction. Two products are formed in this reaction, but one of the products is formed in much greater amounts than the other. (i) Draw the structure of both possible addition products of this reaction.[1] product 1 product 2 H032/01 [2]OCR 2014<br />
(ii) State and explain which of the two possible products will be formed in greater amounts. Include a diagram of the intermediate in the mechanism of this reaction in your answer. 19(iii) 4.125 g of compound D is reacted with an excess of hydrogen chloride. The mixtureof products contains 95% by mass of one product and 5% by mass of the other product. Calculate the mass of each product formed. SPECIMEN [2] [2]OCR 2014 H032/01 Turn over<br />
20 (c) Analysis of one of the compounds A to E is shown below. Percentage composition by mass: C, 78.94%; H, 10.53%; O 10.53% Infrared spectrum: Use this information to identify the compound. Explain your reasoning, referring to all the evidence provided. SPECIMEN[3]OCR 2014 H032/01<br />
21 BLANK PAGE SPECIMENOCR 2014 H032/01 Turn over<br />

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H032/01 Question 22 (i) (a) (b) (iii) Amount of D that reacts Mark Scheme June 20XX ALLOW 4-methyl-3-heptanol ALLOW any unambiguous structure or formula ALLOW ECF on the second structure for hydrogen atom errors if candidate tries to convert to a displayed/structural formula, but the carbon skeleton must be correct If both carbocations are drawn, only one needs to be correct to score the mark ALLOW ECF from (i) for correct justification of product formed in greater amount based on incorrect structures (ii) correct structure of either possible carbocation (cid:57) (cid:57) 1 2 Marks Answer Guidance 4-methylheptan-3-ol (cid:57) SPECIMEN intermediate shown (cid:57) the tertiary halogenoalkane (which will be labelled as either product 1 or product 2) is identified as the one formed in greater amountsbecause the carbocation more stable on C3 than C2 owtte (cid:57) M(D: C7H16O) = 110 (g mol1) AND n(C7H16O) = 4.125 110 Masses of two products formed M(product: C7H17OCl) = 146.5 (g mol1) AND Mass of 95% product = 0.0375 (cid:117) 95 100 AND = 0.0375 (mol) (cid:57) (cid:117) 146.5 = 5.22 g 11 P M T<br />
H032/01 Question (c) June 20XX Mark Scheme Marks Answer Mass of 5% product = 0.0375 (cid:117) 5 100 Guidance (cid:117) 5.49 = 0.27 g (cid:117) 5.49 = 5.22 g (cid:117) 146.5 = 0.27 g (cid:57) ALLOW Mass of both products = 0.0375 (cid:117) 146.5 = 5.49 g Mass of 95% product = 95 100 Mass of 5% product = 5 100 ALLOW product 1 and product 2 if linked to correct mass given labelling in (i) and reasoning in (ii) (ALLOW ECF from (ii)). ALLOW 32003600 cm(cid:16)1 IGNORE references to the peak at ~2900 for CH ALLOW annotation of the spectrum to identify the bond responsible for the peak instead of quoting the wavenumber Conclusion may also follow from empirical formula followed by IR data. SPECIMEN (broad) peak at 33003600 (cm(cid:16)1) for OH (therefore A or C) (cid:57) molar ratio: C : H : O 78.94 12.0 10 : 16 : 1 OR C10H16O, therefore C (cid:57) OR 6.58 : 10.53 : 0.658 (cid:57) : 10.53 16.0 : 10.53 1.0 Total 10 12 P M T<br />

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