OCR A AS Jun 2016 Paper 2 Q3 (with explained solutions) |

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OCR A AS Jun 2016 Paper 2 Q3

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3 Compound A is an alkene. H3C CH3 H compound A CH3 (a) The C=C bond in a molecule of compound A has restricted rotation because it comprises abond and abond. (i) Describe one difference between thebond and thebond.[1] (ii) Explain why compound A does not have E/Z isomers.[1] (iii) A structural isomer of compound A has E/Z isomers. Draw the structure of the Z isomer and then name this isomer. structure of Z isomer name[2]OCR 2016<br />
(b) Compound A can be made from alcohol B by heating with an acid catalyst. Suggest two possible structures for alcohol B. [2]OCR 2016 Turn over<br />
(c)* Compound A reacts with hydrogen bromide to form a mixture of two different organic products. H3C CH3 H compound A CH3 Give the structures of the two possible organic products of the reaction. Outline the mechanism, using the curly arrow model, for the formation of one of the organic products from compound A. Explain which of the two organic products is more likely to be formed.[6]OCR 2016<br />
BLANK PAGE PLEASE DO NOT WRITE ON THIS PAGE Turn over for the next questionOCR 2016 Turn over<br />

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H032/02 Mark scheme June 2016 Marks Question 3 (a) (i) Any one from: Answerbond is between bonding atoms/nuclei ANDbond is above and below the bonding atoms/nucleibond has direct/head-on overlap of orbitals ANDbond has sideways overlapbond has a lower bond enthalpy / is weaker than abondbond has electron density between bonding atoms ANDbond has electron density above and below bonding atoms(ii) One carbon atom (in double bond) is attached to two groups which are identical/the sameAO element AO1.1 Guidance IGNORE the length of thebond andbond IGNORE the type of orbital forbond AO1.1 ALLOWOne carbon atom in (double bond) is not attached to (two) different groups / groups of atomsRight-hand carbon is attached to two groups that are the same/two methyl groups.Two groups are the same on right-hand sideThree groups are the same (on the double bond) DO NOT ALLOWTwo groups on the same side of the double bond Must be right-hand side; Same side could be top or bottom)Functional groups OR molecules for groups<br />
H032/02 Question (iii) (b) Answer Marks (Z-)pent-2-eneAO element AO2.1 AO2.5 AO3.12 10 Mark scheme June 2016 Guidance Mark Independently ALLOW correct structural OR displayed OR skeletal formulae OR a combination of above as long as unambiguous ALLOW C2H5 for CH2CH3 IGNORE connectivity of alkyl groups BUT .DO NOT ALLOW CH3CH2 DO NOT ALLOW trans-pent-2-ene ALLOW correct structural OR displayed OR skeletal formulae OR a combination of above as long as unambiguous ALLOW any vertical bond to OH, e.g. ALLOW OH OR OH | | DO NOT ALLOW OH<br />
H032/02 Question (c) Mark scheme June 2016 Marks Answer Please refer to marking instructions on page 4 of mark scheme for guidance on how to mark this question. Level 3 (56 marks) A comprehensive description with all three scientific points explained thoroughly. There is a well-developed and detailed description of the mechanism, including correct structures, accurately drawn curly arrows and using charges and dipoles consistently. Candidates compare tertiary and secondary carbocation stability to justify major product. Level 2 (34 marks) Attempts to describe all three scientific points but explanations may be incomplete. OR Explains two scientific points thoroughly with no omissions. The description has some structures with reasonably accurate curly arrows and some charges and dipoles identified. Level 1 (12 marks) A simple description based on at least two of the main scientific points OR Explains one scientific point thoroughly with few omissions. The description is communicated in an unstructured way, including some use of curly arrows, charges or dipoles. AO element AO1.22 AO2.52 11 Guidance Throughout: ALLOW correct structural OR displayed OR skeletal formulae OR a combination of above if unambiguous Indicative scientific points 1. Two possible products of reaction CH3C(CH3)BrCH2CH3 CH3CHBrCH(CH3)CH3 IGNORE names where correct structures are present 2. Mechanism for formation of either product. Curly arrow from C=C to attack the H atom of the HBr Correct dipole on HBr Curly arrow from HBr bond to Br Carbocation with full positive charge on carbon atom Curly arrow from negative charge on Br or lone pair on Br- to carbon atom with positive charge OR<br />
Answer Marks 0 marks No response worthy of credit. AO element AO3.1 AO3.2 Guidance 3. Major organic product Major product: 2-bromo-2-methylbutane Major product is formed from the most stable carbocation intermediate OR Br is attached to carbon atom with the least hydrogens attached OR the carbon with the most CH3 groups attached OR the H is attached to the carbon atom with most hydrogens attached H032/02 Question Mark scheme June 2016 Total 12 12<br />

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