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OCR A Jan 2011 Paper 4 Q1
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![Answer all the questions. 1 Methylbenzene, C6H5CH3, is an aromatic hydrocarbon and is used widely as a solvent. It is readily nitrated and it can form mono-, di-, or tri-nitromethylbenzenes. (a) 4-Nitromethylbenzene can be formed by the nitration of methylbenzene. Outline the mechanism for the formation of 4-nitromethylbenzene from methylbenzene using NO2 + as the electrophile. (b) There are six possible structural isomers of CH3C6H3(NO2)2 that are dinitromethylbenzenes. Four of the isomers are shown below. Draw the structures of the other two isomers in the boxes provided. CH3 CH3 CH3 [4] NO2 NO2 isomer 1 CH3 NO2 NO2 isomer 4OCR 2011 O2N NO2 O2N NO2 isomer 2 isomer 3 isomer 5 isomer 6 [2]<br />
(c) A research chemist investigated whether dinitromethylbenzenes could be used in the manufacture of fibres. The chemist devised a two-stage synthesis of the condensation polymer below, starting from one of the isomers in part (b). N H N H N H N H For the first stage of the synthesis,Which of the isomers 1, 2, 3 or 4 could be used? Identify the product formed and state suitable reagents.Write an equation. For the second stage of the synthesis, Suggest an organic compound that could react with the organic product from the first stage to form the polymer. State the type of condensation polymer formed.[6]OCR 2011 [Total: 12] Turn over<br />](https://learnah.org/wp-content/uploads/2018/08/OCR_4_QP_JAN11_question_1-0.jpg)
![F324 Mark Scheme January 2011 Question Answer 1 (a) CH3 CH3 CH3 ALLOW Kekul structures throughout Mark Guidance ANNOTATIONS MUST BE USED ALLOW skeletal CH3 ALLOW +NO2 OR NO2 ALLOW 1st curly arrow from the ring OR from within the ring to any part of the NO2 DO NOT ALLOW intermediate with broken ring less than halfway down: + including the + charge CH3 Horseshoe must have open end towards NO2 ALLOW Kekul mechanism: NO2 CH3 CH3 CH3 NO2 curly arrow from ring + to NO2 NO2 correct intermediate curly arrow from CH bond back to reform ring H+ NO2 correct products (cid:57) (cid:57) (cid:57) (cid:57) 1 mark for intermediate 1 mark for curly arrow H+ + NO2 + NO2 NO2 ALLOW double bonds shown in other Kekul arrangement IF CH3 has been omitted completely (ie benzene shown), DO NOT AWARD intermediate mark OR products mark (max 2) IF NO2 is shown in incorrect position in intermediate or product, DO NOT AWARD intermediate mark but award other marks (max 3)<br />
F324 Question Answer 1 (b ) O2N CH3 CH3 NO2 NO2 (cid:57) (cid:57) 1st stage isomer: isomer 3 (cid:57) product: CH3 (cid:57) NH2 H2N reagents: Sn AND (conc) HCl (cid:57) equation: CH3 (c) O2N Mark Scheme January 2011 Mark Guidance ALLOW any correct unambiguous structures ALLOW NO2 Note: connectivity is NOT being assessed in this part NO2 ANNOTATIONS MUST BE USED ALLOW structure of isomer 3 shown separately OR in equation ALLOW structure of product shown separately OR in equation ALLOW correct name (3,5-diaminomethylbenzene) IGNORE incorrect name DO NOT ALLOW CH3C6H3(NH2)2 ALLOW Zn + HCl/H2 + metal catalyst/LiAlH4/Na in ethanol IGNORE NaBH4 ALLOW Sn and HCl followed by NaOH DO NOT ALLOW Sn and HCl and NaOH IF isomer 3 OR product are given in equation but not shown previously then credit here Also credit reagents here if shown (eg above arrow) ALLOW correct structural OR displayed OR skeletal formula ALLOW combination of formulae as long as unambiguous CH3 12 [H] NO2 H2N 4 H2O NH2 (cid:57)<br />
Mark Scheme January 2011 Mark Guidance HO OH DO NOT ALLOW molecular formula ALLOW name of compound: propanedioic acid OR propane-1,3-dioic acid ALLOW absence of e after propan ALLOW acyl dichloride: ClOCCH2COCl ALLOW cyclic acid anhydride of propanedioic acid: CH2 ALLOW Nylon or Kevlar DO NOT ALLOW polypeptide DO NOT ALLOW amide F324 Question Answer (i) 2nd stage (c) organic compound: HOOCCH2COOH (cid:57) type of polymer: polyamide (cid:57) Total 12<br />](https://learnah.org/wp-content/uploads/2018/08/OCR_4_QP_JAN11_answer_1-0.jpg)
