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CIE Jun 2013 v2 Paper 4 Q3
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![Examiners For Use (a) Describe the reagents and conditions required to form a nitro compound from the following. (i) methylbenzene CH3 OH(ii) phenol[3] (b) Draw the structure of the intermediate organic ion formed during the nitration of benzene. (c) In the box over the arrow below, write the reagents needed to convert nitrobenzene into phenylamine. NO2 NH2 [1] [1]UCLES 2013 9701/42/M/J/13<br />
Examiners For Use (d) Phenylamine can be converted into the organic compounds A and B. (i) Suggest the structural formulae of A and B in the boxes below. (ii) Suggest suitable reagents and conditions for step 1, and write them in the box over the arrow. NH2 Br2(aq) step 1 N+ ClO Na+ [3] (e) When phenylamine is treated with propanoyl chloride a white crystalline compound, C, C9H11NO, is formed. (i) Name the functional group formed in this reaction.(ii) Calculate the percentage by mass of nitrogen in C. (iii) Draw the structural formula of C. percentage =%UCLES 2013 9701/42/M/J/13 [3] [Total: 11] [Turn over<br />](https://learnah.org/wp-content/uploads/2018/08/June_2013_v2_QP_-_Paper_4_CIE_Chemistry_A-level_3-0.jpg)
![Page 5 Mark Scheme Syllabus Paper GCE A LEVELMay/June 2013 9701 42 (c) (i) CH3CO2Na + HClCH3CO2H + NaCl (ii) CH3CO2H + NaOHCH3CO2Na + H2O PMT [1] [1] [2] (d) e.g. hydrolysis of esters RCO2R' (+ H2O)RCO2H + R'OH or its reverse or hydrolysis of amides: RCONH2 (+ H3O )RCO2H + NH4 hydrolysis of nitriles: RCN (+ H3O + H2O)RCO2H + NH4 nitration of benzene (or any arene): C6H6 + HNO3C6H5NO2 (+ H2O) dehydration of alcohols, e.g. : CH3CH(OH)CH3CH3CH=CH2 + H2O (or the reverse) halogenation of ketones, e.g. : CH3COCH3 + X2CH3COCH2X (+ HX) [3] [Total: 17] (a) (i) HNO3 + H2SO4 conc (both acids) and 30 C < T < 60 C or warm (ii) dilute HNO3 or HNO3(aq) and room temp. (allow T30 C) (b) (allow intermediate from methylbenzene) NO (c) Sn/tin (or SnCl 2, Fe) + HCl (NOT H2SO4 or H , Zn, or LiAlH4.) [1] [1] [1] [3] [1] [1] [1] [1]Cambridge International Examinations 2013<br />
Page 6 Mark Scheme Syllabus Paper GCE A LEVELMay/June 2013 9701 42 PMT OH or -ONa or -O NOT -NaO (d) (i) or NH NH Br Br Br (ii) NaNO2 + HCl or H2SO4 or H or HNO2 T10 (e) (i) amide (ii) M r = 108+11+14+16 = 149 %N = (14 x 100)/149 = 9.4% (iii) NHCOC [1] + [1] [1] [1] [4 max 3] [1] [1] [1] [3] [Total: 11] 4. (a) (i) Many electrons of similar energy in a valence-shell orbital or successive ionisation energies rise steadily (no big jumps) or ability to form bonds with ligands can stabilise very low or very high oxidation states or 4s + 3d orbitals/shells/energy levels have similar / same energies (ii) VO2 : +5 CrF6 : +4 2 MnO4 : +6 2 [1] [31] [4]Cambridge International Examinations 2013<br />](https://learnah.org/wp-content/uploads/2018/08/June_2013_v2_MS_-_Paper_4_CIE_Chemistry_A-level_3-0.jpg)
