JF Annual Paper II 2002 Model Answers

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1 JF Annual Paper II 2002 Model Answers Q1. yclohexane conformation chair more stable than the boat conformation (not answer d). For substituted cyclohexanes axial position is less stable than equatorial position due to 1,3-diaxial interactions (not answer b or e). Axial positions Equatorial positions Axial and equatorial positions Answer is (a) or (c). (a) is the trans isomer and (c) is the cis isomer.. Q2. Reaction mechanism (a) is wrong, reaction mechanism (b) is wrong, reaction mechanism (c) is correct.. Q3. Remember experiment 7 of the organic labs! When methylbenzoate is refluxed with sodium hydroxide for half an hour and then treated with l to give benzoic acid. X = methylbenzoate. 3 Na + Methylbenzoate Benzoic acid White solid

2 Q4. The Iodoform reaction is a qualitative test for methyl ketone structural unit. R 3 R 3 Methyl ketone unit Alcohol unit produced form oxidation of a methyl ketone The ketone is oxidised to an alcohol therefore the alcohol must have the above structural unit. Therefore the correct answer is 2-hexanol hexanol Q 5. 2-ethyl-1-pentene was hydrogenated to give compound X. X =? ethyl-1-pentene 2 3 ompound X Name compound X Longest chain contains 6 carbon, therefore exane and the 3 group is attached to carbon 3. ence the name is 3-methylhexane methyl-2-hexene ompound X ompound X = 3-methylhexane

3 Q6. Tertiary haloalkanes undergo elimination type reactions under strong basic conditions (answer has to be either d or e). Apply Saytzeff s Rule formation of the most highly substituted alkene. l chloro-2-methylbutane 2-methyl-2-butene Q7. X + hydroxylamine ydroxylamine 2 N R R + 2 N N R R R R N ydroxylamines condense with aldehydes and ketones to form imine products. X = ketone Q8. zonolysis: zidative cleavage of alkenes to carbonyl compounds 3 Reduction -[] + Single product formed therefore starting material must by symmetrical. Product is propanal. Therefore starting material is 3-hexene.

4 hexene Reduction -[] Propanal Q9. 2Na(s) Na Na Na + 3 When sodium metal is dissolved in methanol sodium methoxide is formed and is a white solid when the excess methanol is evaporated off. This solid is dissolved in water to give the hydroxide, when evaporated to dryness a white solid is obtained which is sodium hydroxide. Answer (b) Q10. (a) Z isomer E isomer (b) 3 l 3 l (c) l 2 (d) Br Br Br Br

5 (e) 3 2 Br Br chloropropene does not exhibit geometrical (E/Z) isomerism. Q11. Para 3 Para Meta Meta rtho 3 group is electron donating, activating group. Activating groups are ortho and para directors therefore, the nitration of toluene will take place at the ortho and para positions. Q cm 3 of 0.08M 2 2 contains? number of moles M = molarity = number of moles in 1 litre Therefore 100 cm 3 contains 100 cm M/1000 cm 3 = moles 2 2 From the balanced equation above we know that: 2 moles of 2 2 produces 1 mole of 2 or 1 mole 2 2 produces ½ moles of 2 so therefore moles 2 2 produces moles 2 Use the ideal gas law to calculate the volume PV = nrt P = 1 atm R = L atm K -1 mol -1 V =? T = 298 K n = moles

6 V = ( mol)( L atm K -1 mol -1 )(298 K) (1 atm) V = L or V = 97.8 cm 3 Answer (b) Q13. Write balanced equations for the combustion of each of the organic compounds. The answer will be the one that has 2 moles of 2 and 2 moles of 2. (a) 3 + 2½ (b) (c) (d ) (e) ½ Q14. Pb + 2 Pb 2 f =? Given 3Pb Pb 3 4 = kjmol -1 3Pb 2 Pb = kjmol -1 Therefore Pb Pb 2 = kjmol -1 3Pb Pb 3 4 = kjmol -1 Pb Pb 2 = kjmol -1 3Pb Pb 2 = kjmol -1 Pb + 2 Pb 2 = kjmol -1 Answer (e) Q15. g(l) g(g) = 61.3 kjmol -1 S = = 99 Jmol -1 K -1 Boiling point = / S = Jmol -1 /99 Jmol -1 K -1 Boiling point = 619 K Answer (e)

7 Q16. All statements are wrong. (a) ccp. Ratio = 0.74; bcp ratio = 0.68 (b) diamond is tetrahedrally coordinated (cccp coordination number is 12) (c) simple cubic is very inefficient packing ratio = 0.52 (d) entropy increases on melting (e) in sl each s has 8 nearest neighbours Answer (e) Q17. (i) Rate depends on [A] not [A] 2 as required for second order kinetics. (ii) Arrhenius equation k = Ae -Ea/RT R = kjmol -1 (iii) the first step is rate determining. Therefore, first order with respect to A (iv) this statement is wrong as would show second order kinetics. Answer (e) Q18. T = K f m (m is molality) Therefore T = = K K = 0.13 T = = 5.37 Q19. p = -log[ 3 + ] Therefore [ 3 + ] = inverse Log of p p = 8.2 [ 3 + ] = inverse Log of 8.2 = M

8 Q20. The p of 1M is Ka =? is a weak acid p = -log[ 3 + ] [ 3 + ] = Ka = [ 3 + ][A - ] [A] onc. A 3 + A - Initial hange -x +x +x Equilibrium 1-x +x +x Ka = [x][x] x 2 Ka = x 2 [1-x] 1 x = [ 3 + ] = Ka = (0.0135) 2 = Answer (a) Q21. E cell = E cathode - E anode E cell = 1.24 V Ag + /Ag E = 0.80 V athode Reduction occurs IL RIG xidation is loss of electrons Reduction is gain of electrons Anode xidation ccurs E cathode = 0.80 V 1.24 V = 0.80 V - E anode E anode = V Answer (b)

9 Q22. KAul 4 Au Au (3+) Au (0) 1 mole of Au 3 moles of electrons moles of Au moles of electrons = moles of electrons Faraday s Law Number of moles = urrent (A) time (s) Faraday s constant = t t = 3178 seconds t = 53 mins n = It F Q23. de Broglie λ = h = h mc p p = momentum Kinetic Energy = p 2 2m p m p = rest mass of proton = kg (value given in table of constants at the back of exam paper). Therefore: J = p 2 2m p p = ( ) = kg m s -1 λ = h p λ = λ =

10 Q24. Sulfate ionb S 4 2- S x x x S xx x Formal charge(q) = no.of valence electrons(v) (no. of unshared electrons(l) + ½ no. of shared electrons(s)) Therefore q = v (l + ½ s) q S = 6 (0 + ½ 12) = 0 (formal charge sulphur) q = 6 (4 + ½ 4) = 0 (formal charge doubly bonded oxygen) q = 6 (6 + ½ 2) = -1 (formal charge singly bonded oxygen) Answer (a) Q25. Ferric ion Fe 3+ there are 5 electrons in the outer 3d shell 3d 5 (a) Spin = + ½ + ½ + ½ + ½ + ½ = 2.5 if it gains one electron then it becomes 3d 6 then spin = + ½ - ½ + ½ + ½ + ½ + ½ = 2 The ion is less paramagnetic. True (b) The outer shell contains 5 unpaired electrons True (c) There are no electrons in the 4s shell False (d) reduction is gain of electron so 3d 5 3d 6 True (e) spin = 2.5 is maximum spin for 3d True Q26. Xenon tetrafluoride XeF 4. F F x. Xe F. x x x F.... Two lone pair lobes which minimise their repulsion by occupying axial positions. Since the 6 electron-pair lobes point to the corners of an octahedron the bonding pairs must be equatorial. So the molecular shape is square planar.

11 Answer (b) F F.. F Xe F. Square planar Q g mass of MS Molar Mass (MM) of MS = M g mass of M Molar Mass (MM) of M = M +16 number of moles of MS = number of moles M = M + 32 M M + 16(1.164) = 0.972M + 32(0.972) 1.164M 0.972M = M = M = 65 gmol -1 M = Zn Q28. Mg + 2l Mgl 2 (aq) Na 2 3 Mg (s) + 2 (g) Mg 3 (s) + 2Nal(aq) Answer (b) Q29. E has two oxidation states II and IV therefore maybe in Group 14. Reacts with acid to give 2 (g) therefore metal. Answer (e) Q30. Fe P 4 2FeP MM Fe 2 3 = 2(55.8) + 3(16) = gmol -1 Number of moles Fe 2 3 = 1g gmol -1 = 6.26 mmol Number of moles 3 P 4 = 20 ml 0.1/1000 = 2 mmol

12 Fe 2 3 is in excess and 3 P 4 is the limiting reagent therefore 2 mmol of FeP 4 is maximum possible number of moles that could be pproduced. Answer (b) Q ml 0.1/1000 = number of moles of + = 3.6 mmol 24 ml 0.075/1000 = number of moles of - = 1.8 mmol 3.6 : 1.8 = 2:1 Reaction ratio = 2:1 Q32. (i) highest oxidation number +5 lowest 3 This statement is false (ii) all non-metal oxides are acidic This statement is true (iii) N triple bond N has a bond order of 3 This statement is false (iv) 2N 2 N 2 4 This statement is true (v) N 2 boiling point 195 o This statement is true

e.g. propan-2-ol ethane-1,1-diol propane-1,2,3-triol H H

e.g. propan-2-ol ethane-1,1-diol propane-1,2,3-triol H H Alcohols General The functional group is - The homologous series has general formula n 2n+1 The names end in ol, with a number if needed to indicate where on the carbon skeleton the group is located. When