AQA AS Jun 2016 Paper 2 Q7 (with explained solutions) |

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AQA AS Jun 2016 Paper 2 Q7

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16The alkene 3-methylpent-2-ene (CH3CH=C(CH3)CH2CH3) reacts with hydrogen bromide to form a mixture of 3-bromo-3-methylpentane and 2-bromo-3-methylpentane. 0 7 . 1 The alkene 3-methylpent-2-ene (CH3CH=C(CH3)CH2CH3) exists as E and Z stereoisomers. Draw the structure of Z-3-methylpent-2-ene. [1 mark] 0 7 . 2 Name and outline the mechanism for the formation of 3-bromo-3-methylpentane from this reaction of 3-methylpent-2-ene with hydrogen bromide. Explain why more 3-bromo-3-methylpentane is formed in this reaction than 2-bromo-3-methylpentane. [7 marks] *16* M/JUN16/7404/2<br />
17 Turn over for the next questionDO NOT WRITE ON THIS PAGE ANSWER IN THE SPACES PROVIDED *17* Turn overM/JUN16/7404/2<br />

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MARK SCHEMEAS-LEVEL CHEMISTRY 7404/2JUNE 2016 Marking Guidance Mark Comments Must show all 4 groups bonded to C=C Allow CH3 for methyl group; allow C2H5 for ethyl group Allow correct structure of the style Allow correct skeletal structure 20 of 24 Question 07.1<br />
07.2 M1 electrophilic addition NB the arrows here are double-headed M2 must show an arrow from the double bond towards the H atom of the H-Br molecule M3 must show the breaking of the H-Br bond M4 is for the structure of the tertiary carbocation M5 must show an arrow from the lone pair of electrons on the negatively charged bromide ion towards the positively charged atom (of either a secondary or) of a tertiary carbocation M6 3-bromo-3-methylpentane is formed from 3y carbocation OR 2-bromo-3-methylpentane is formed from 2y carbocation M7 3y carbocation more stable than 2y MARK SCHEMEAS-LEVEL CHEMISTRY 7404/2JUNE 2016 1 1 1 M2-M5 Penalise one mark from their total if half-headed arrows are used M2 Ignore partial negative charge on the double bond M3 Penalise incorrect partial charges on H-Br bond and penalise formal charges Penalise M4 if there is a bond drawn to the positive charge Penalise only once in any part of the mechanism for a line and two dots to show a bond Max 3 of any 4 marks (M2-5) for wrong organic reactant or wrong organic product (if shown) or secondary carbocation Max 2 of any 4 marks in the mechanism for use of bromine Do not penalise the correct use of sticks For M5, credit attack on a partially positively charged carbocation structure but penalise M4 M6 is high demand and must refer to product being formed from/via correct class of carbocation M7 is high demand and must be clear answer refers to stability of carbocations (intermediates) not products Candidate that states that products are carbocations would lose M6 and M7 M6,7 allow carbonium ion in place of carbocation; or a description of carbocation in terms of alkyl groups/ number of 21 of 24<br />
MARK SCHEMEAS-LEVEL CHEMISTRY 7404/2JUNE 2016 carbon atoms joined to a positive C When asked to outline a mechanism, candidates are expected to draw a mechanism with curly arrows (specification 3.3.1.2). On this occasion only we would allow a detailed description as shown. M2 must the describe the movement of a pair of electrons / curly arrow from the C=C towards the H atom of the H-Br molecule M3 must describe the breaking of the H-Br bond with the bonding pair of electrons moving to the Br / curly arrow from H-Br bond to Br M4 is for the structure of the tertiary carbocation (i.e. positive C bonded to one methyl and two ethyl groups) M5 must describe the movement of a pair of electrons from the Br ion to the positive C atom of the carbocation / curly arrow from the lone pair of electrons on the negatively charged bromide ion towards the positively charged C atom (of either a secondary or) of a tertiary carbocation 22 of 24<br />

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