A-LevelChemistryHydrocarbons9701/41 · May/June 2020 · Q514 Marks

(a) Benzene reacts with bromine in the presence of an aluminium bromide catalyst, AlBr3, to form bromobenzene. This is a substitution reaction. No addition reaction takes place.

(a)(i)Explain why no addition reaction takes place.[1]
(a)(ii)AlBr3 reacts with bromine to generate an electrophile, Br⁺. Draw the mechanism of the reaction between benzene and Br+ ions. Include all relevant arrows and charges. [Figure for drawing the mechanism][3]
(a)(iii)Write an equation to show how the AlBr3 catalyst is reformed.[1]
(b)Suggest why bromination of phenol occurs more readily than bromination of benzene.[2]
(c)(i)There are four different carbocations with the same formula, C4H9⁺. One structure is given in the table. Suggest the structural formulae of the three other carbocations. [Table with structure 1: CH3CH2CH2CH2+] structure 2 structure 3 structure 4[3]
(c)(ii)Benzene reacts with each of these carbocations in separate Friedel-Crafts alkylation reactions. In each reaction an organic compound with formula C10H14 is formed. The number of peaks observed in the carbon-13 NMR spectrum of each compound is given. Suggest the structures for the three other compounds. [Table with a drawn structure for 1-phenylbutane with 8 peaks, and three blank spaces with 6, 7, and 8 peaks respectively for C-13 NMR][4]

This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.

Mark scheme

Marking points from the official mark scheme (May/June 2020). The mark or mark type is shown in brackets.

[1] The substitution product is stabilised by delocalisation of (6)π-electrons OR The addition product is not stabilised by delocalisation of (6)π-electrons [1]
[3] [Diagram showing curly arrow from benzene to Br+, carbocation intermediate, curly arrow from C-H bond to reform ring, and H+ product] • first curly arrow • intermediate • 2nd curly arrow, product and H+ formed / lost
[1] AlBr4 + H+ → AlBr3 + HBr
[2] lone pair of oxygen is delocalised into the ring
[2] any one from: • phenol has a higher electron density in the ring • phenol can polarise/induce a dipole in Br2
[3] CH3CH2CH+CH3 (CH3)2CHCH2+ (CH3)3C+ Each correct structure = 1 mark
[4] [Diagrams for the 4 isomers of C10H14, each with their correct NMR peak count. The first one is 1-phenylbutane (8 peaks), then 2-phenylbutane (8 peaks), isobutylbenzene (7 peaks), tert-butylbenzene (6 peaks)] Two correct organic products = 1 mark three correct organic products = 2 marks all products linked correctly to NMR = 2 marks

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