JJC H1 Chemistry 8872 H1 Prelim Paper 2 Ans

JJC Chemistry H1 P2 Answers 1. (a) (i) Amount of Cl2 in 1 cylinder = 0 . 71 170000000 / 5730 = 4.18 x 102 mol [1] (ii) Conc. of Cl2 in g dm–3 = mg g dm mg 1000 1 11000 3 × × = 1 Conc. of Cl2 in mol dm–3 = 0 . 71 1 = 0.0141 mol dm–3 [1] (a) (iii) Volume of factory room = 3 3 3 3 10 1 125 m dmm −×× = 25 000 dm3 Conc. of Cl2 in the factory = 25000 10 18 . 42× = 0.0167 mol dm–3 0.0167 mol dm–3 > 0.0141 mol dm–3 / lethal conc [2] (b) (i) C O Cl Cl 120o [1] (ii) sp2 [1] (iii) Phosgene contains both σ bonds and π bonds. Explain the terms in italics, drawing diagrams to show these bonds clearly. State the number of σ bonds and π bonds in phosgene. A sigma bond (σ bond) is formed by head-on overlap of orbitals. A pi bond (π bond) is formed by side-way overlap of orbitals. Phosgene has 3 σ bonds and 1 π bond. (iv) Both phosgene and chlorine have simple covalent / molecular structure Higher amount of energy is required to overcome the stronger permanent dipole – permanent dipole attraction / van der Waals’ forces of attraction/ intermolecular forces between phosgene molecules than weaker in induced dipole – induced dipole attraction / van der Waals’ for ces of attraction/ intermolecular •• π bond σ bond Page 1 of 12

forces between chlorine molecules. OR Phosgene has a larger size of electron cloud / no. of electrons than chlorine . Higher amount of energy is stronger permanent dipole – permanent dipole attracti on / van der Waals’ forces of attraction/ intermolecular forces between phosgene molecules than weaker in induced dipole – induced dipole attraction / van der Waals’ forces of attraction/ intermolecular forces between chlorine molecules. Thus phosgene has a higher boiling point than chlorine. [2] [Total: 11] 2. (a) The 1st IE of an element E is the energy required to remove one mole of electrons from one mole of gaseous E atoms. [1] (b) [1] (c) Explain the difference between the first ionisation energy of (i) carbon and nitrogen 6C: 1s2 2s2 2p2 7N: 1s2 2s2 2p3 N has a smaller atomic radius than C and N has a larger nuclear charge than C while shielding effect by the inner shell electrons is relatively constant for both elements. Hence 1 st IE of N is higher than that of C. [1] (ii) nitrogen and oxygen proton number First ionisation energy (kJ mol−1) 5 Li Be B C N O F Ne Li Be B C N O F Ne 0 10 Page 2 of 12

N: 1s2 2s2 2p3 O: 1s2 2s2 2p4 Mutual repulsion between the paired 2p electrons in O makes it easier to remove one of the paired electrons compared to removing the unpaired 2p electron of P which does not experience such repulsion. Hence 1st IE of O is lower than that of N. [1] [Total: 4] (i) [acetylsalicylic acid]= 1000 0 . 25 200 . 01000 20 × = 0.160 mol dm–3 [1]3. (a) (ii) [H+] = 10-2.8 = 1.58 × 10-3 mol dm-3 [H+] << or ≠ [acetylsalicylic acid] OR Degree of acid dissociation = 0.160 10 8 1.53−× = 9.88 × 10-3 ∴acetylsalicylic acid is a weak acid as it only dissociates/ionises partially in water. [2] (iii) 87 2 87 2 2 [] [ ] [] a CHC O HK CHOC OH −+ = OR ] [ ] ][ [ RCOOH H RCOOKa + − = 32(1.58 10 ) 0.160 −×= = 1.56 × 10-5 mol dm-3 OR 3 2 3 10 58 . 1 160 . 0 ) 10 58 . 1 ( − − × − ×= = 1.59 × 10-5 mol dm-3 [2] 3. (b) (i) Ka increases increasing number of Cl substituents Both CH2ClCOOH and CHC l2COOH has electron-withdrawing C l atom which disperse the negative charge on O atom on the RCOO−/ carboxylate anion, making CH2ClCOO− and CHCl2COO− more stable than CH3COO−. CHCl2COOH has one more electron-withdrawing C l atom than CH2ClCOOH, dispersing the negative charge on O atom to a larger extend / more, making CHCl2COO− more stable than CH2ClCOO [2] Page 3 of 12

(ii) Benzoic acid is more acidic than ethanoic acid The negative charge on the COO − group of the benzoic acid can be delocalised into the π system of the benzene ring / benzene ring , dispersing the negative charge on the COO − anion, making the benzoate anion more stable than CH3COO− . [2] [Total: 9] 4. 4. (a) (i) Draw the structures of compounds A to C in the boxes provided. A B NaOH (aq) heat under reflux CH=CHCH2Cl Stage II Acidified K2Cr2O7 COCH2COOH 2,4-dinitrophenylhydrazine C Stage III CCH2CH2OH CH2NH2 O H Stage IV Cinnamyl chloride Stage I CCH2COOH CN O H heat under reflux CH=CHCH2OH CH(OH)CH2CH2OH CCH2COOH N N H O2N NO2 [1m] [1m] [1m] Page 4 of 12

(ii) State the type of reaction and r eagents and conditions required for Stages II to IV. Stage Reagents and conditions Type of Reaction II H2O (g) / steam H3PO4 cat. , 300oC , 65 atm Addition /Hydration III HCN, trace amt of NaCN/KCN/base Addition IV LiAlH4, dry ether Reduction [1m] each [6] (iii) Write a balanced equation for the reaction in Stage IV. CCH2COOH CN O H CCH2CH2OH CH2NH2 O H + 4[H] [1m] [10] + H2O 8[H] 4. (b) (i) and W X H2SO4(aq), heat (acid hydrolysis) + CH3CH2OH + (CH3)3COH COOCH2CH3 COOC(CH3)3 COOH COOH Test : Add H2SO4(aq) to each of the samples and heat. To the resulting hot mixture, add a few drops of acidified KMnO4(aq) / K2Cr2O7 and heat. Observations: W will decolourise purple KMnO 4 / orange K 2Cr2O7 turns green but X will not. [3] COOCH2CH3 COOC(CH3)3 Cannot be oxidised Page 5 of 12

(ii) and Y Z CH3 CH3 acidified KMnO4(aq), heat (oxidation) COOH CH3 O COOH COOH CH3 Test: Add acidified KMnO4(aq) to each of the samples and heat. To the resulting mixture, add Brady's reagent /2,4- dinitrophenylhydrazine. Observations: Y will decolourise purple KMnO 4 and give an orange ppt with the Brady's reagent. Z will also decolourise purple KMnO 4 but will not form an orange ppt with Brady's reagent. [3] [Total : 16] CH3 CH3 can undergo condensation reaction with 2,4- DNPH SECTION B (40 MARKS) Answer two of the three questions in this section on separate paper. 5. (a) (i) Standard enthalpy change of combusti on of a substance is the enthalpy change/ energy change when one mole of substance is completely burnt in oxygen under standard conditions. CH3CH3 (g) + 2 7 O2 (g) Æ 2CO2 (g) + 3 H2O (l) [2] (ii) Heat evolved = m c ΔT = 500 x 4.2 x 40 = 84000J = 84 kJ ΔHc (ethane) = − 84 / 0 . 30 5 . 2 = − 1008 kJ mol–1 [2] (iii) Heat loss to the surroundings [1] Page 6 of 12

(b) (i) ΔHc θ (ethane) = ) ( ) tan (productsBE ts reac BE∑−∑ = Σ[BE (C─C) +6BE(C─H)+ 2 7 BE(O=O)] – Σ[4BE(C=O) + 6 BE(O─H)] = [(350 + 6(410) + 2 7 (496)] – [4(740)+6(460)] = –1174 kJ mol–1 [2] (ii) 2 C (graphite) + 3 H2 (g) CH3CH3 (g) 2 C O 2 (g) + 3 H2O(l) ΔHf θ (ethane) = 2(–394) + 3(–286) – (–1174) = – 472 kJ mol–1 [2] (c) (i) Deduce the identities of compounds P to T , explaining the chemistry of the reactions wherever possible. P (C8H10) Cl2 Q (C8H9Cl) hot alc KOH CH=CH2 hot aq. NaOH S warm aq. alkaline I2 pale yellow pptCl2 R (C8H9Cl) hot acidified KMnO 4 T (C7H5ClO2) hot aq. NaOH no rxn 2Δ cH θ (C) = 2(–394) 3ΔHc θ (H2) = 3(–286) ΔHc θ (ethane) = –1174 Page 7 of 12

• M.F C8H10 : C/H ratio ≈ 1 ÆP has a benzene ring • Q reacts with hot alcoholic KOH : Elimination Q contains halogenoalkane/ alkylhalide group • Q reacts with hot aq. NaOH : Substitution Q contains halogenoalkane/ alkylhalide group S : contains –OH group / hydroxyl group • S reacts with warm aq. I2 : oxidation / positive iodoform test / positive tri-iodomethane test Q contains structural unit • R does not contains halogenoalkane/ a