SRJC H1 CHEM P2 ANSWER

2011 H1 MYE Section A Marker’s Remarks 1 Elements D and E are period 3 elements. (a) D is a white solid with melting point of 317 K. When exposed to air, it reacts spontaneously, producing a white powder of empirical formula D2O5. D2O5 readily undergoes neutralisation with aqueous NaOH to form salt and water. (i) From the observations , deduce an identity of D. Include equations, with state symbols, for any reactions that has occurred. Low melting point hence D must be of simple molecular structure: sulfur or phosphorus. The oxide of D is a white powder with empirical formula D2O5 hence must be phosphorus as sulf ur is yellow and its oxide does not have empirical formula of D2O5. P4 (s) + 5 O2 (g) P4O10 (s) P4O10(s) +12 NaOH(aq) 4 Na3PO4(aq) + 6 H2O(l) (ii) Suggest and account for a pH value for the resultant solution when the oxide of D dissolves in water. P4O10(s) + 6H2O(l) 4 H3PO4(aq) Resulting pH: 2 or 3 [6] (b) The successive ionisation energies of element E is shown below. [2] 1st 2nd 3rd 4th 5th 6th 7th 8th I.E. kJmol-1 999 2265 3331 5071 7008 8487 27107 31719 (i) Deduce the group number in which E belongs. Group VI (ii) Write down the electronic configuration of E in the +4 oxidation state. 1s2 2s2 2p6 3s2 [2] (c) Compare the 1st ionisation energy of E with the element just before it. Explain. E will have a lower 1st I.E. compared to the element just before it. Due to inter-electronic repulsion between the pair of electrons sharing the same orbital, thus lesser amount of energy is required to remove an electron from Element E. [2] Total: [10]

2 Manganese dioxide, MnO2 is a strong oxidising agent and it is also commonly used as a catalyst. (a) On the grid below, sketch and label the energy distribution of gas molecules in the presence of a catalyst. Your sketch should clearly illustrate the effect of catalyst on rate of reaction. [2] (b) The lattice energy of MnO 2 can be calculated based on the energy cycle shown below. Mn2+ (g) + 2 O2− (g) Mn2+ (g) + G + 2e Mn2+ (g) + O2 (g) + 2e Mn+ (g) + O2 (g) + e Mn(g) + O2 (g) Mn(s) + O2 (g) MnO2 (i) Identify the species G, with appropriate state symbol G: O(g) Energy Fraction of molecules BE (O2) 1 (Mn) 2 3 1st + 2nd E.A (O) 4 0 Energy

(ii) Identify the enthalpy changes labeled by the numbers 1 to 4. 1 H 2 1st ionisation of Mn 3 2nd ionisation of Mn 4 lattice energy of MnO2 [3] (c) Potassium manganate (VII), KMnO 4, can be manufactured from MnO 2. During the process, MnO 2 is reacted with concentrated KOH to produce potassium manganate ( VI), K 2MnO4. When a solution of K 2MnO4 is acidified, a disproportionation reaction occurs. This produces a dark brown MnO2 precipitate and a solution of KMnO4. (i) With reference to oxidation states of manganese, explain what is meant by disproportionation and suggest an equation for this reaction. Disproportionation is a reaction where reduction and oxidation of the same element occurs simultaneously. Mn changes its oxidation state from +5 to +4 and + 7 3MnO4 2− + 4H+  2MnO4 − + MnO2 + 2H2O (ii) In alkaline solution, KMnO4 reacts with sodium sulfite, Na2SO3 in a 1:1 ratio. Suggest the final oxidation state of manganese. [Given: SO3 2− + 2OH-  SO4 2− + H2O + 2e] Since MnO4 − ≡ SO3 2−, The number of electrons transferred = 2 The final oxidation state of Mn = 7-2 = +5 [5] Total: [10]

3 H is a halogenoalkane with molecular formula C4H9Cl. When reacted with hot ethanolic NaOH, the product, J is found to display geometrical isomerism. When refluxed with aqueous NaOH, compound K is formed. K undergoes oxidation when reacted with two different sets of reagents and conditions. When K is oxidised using the first s et of reagents and conditions, followed by acidification, propanoic acid is formed. However, when K is oxidised using the second set of reagents and conditions, another product L is formed. L reacts with 2,4 -DNPH but not with Tollens' reagent. (a) Using the information given above, deduce the structures of compounds H, J, K and L and draw the structural formulae of the molecules in the relevant boxes provided in the flowchart. [4] H C H H C H H C H C Cl H H H Compound H hot ethanolic NaOH H C H H C H C H C H H H Compound J hot aqueous NaOH H C H H C H H C H C H H HOH Compound K Oxidation I followed by acidification propanoic acid Oxidation II H C H H C H H C C H H HO Compound L

(b) State the reagents and conditions for Oxidation I and Oxidation II. Oxidation I: alkaline I2(aq), warm Oxidation II: KMnO4(aq)/K2Cr2O7(aq), H2SO4(aq), heat [2] (c) Draw and name the two isomers of compound J. C C H CH3 CH3 H C C CH3 CH3 HH trans-but-2-ene cis-but-2-ene [2] (d) Compound K is found to have a higher boiling point as compared to compound L. With reference to structure and bonding, explain why this is so. Compound K has a simple molecular structure with intermolecular hydrogen bonds while compound L has a simple molecular structure with intermolecular van der Waals' forces of attraction . More energy is needed to break the stronger hydrogen bonds between molecules of K compared to the weaker van der Waals' forces of attraction between molecules of L . Hence, K has a higher boiling point than L. [2] Total: [10]

4 Halogenoalkanes, R X, are important intermediates used to produce many organic compounds. They may be converted into Grignard reagents by reacting them with magnesium in dry ether: RX + Mg R Mg X where R is an alkyl group and X is chlorine, bromine or iodine. (a) (i) With relevant data from data booklet, explain why the reactivity between halogenoalkanes and Mg is in the following order: iodoalkane > bromoalkane > chloroalkane C-Cl bond: 340 kJ mol−1 C-Br bond: 280 kJ mol−1 C-I bond: 240 kJ mol−1 Strength of C-X bond: C-Cl> C-Br > C- I Ease of breakage of the bond: C- I > C-Br > C-Cl Therefore reactivity: iodoalkane > bromoalkane > chloroalkane On reaction with carbonyl compounds, RMgX forms various classes of alcohols. The reaction below shows an example of the formation of a secondary alcohol: Aldehyde + RMgX 1. Dry ether 2. H3O+ secondary alcoholother than methanal C O R' H C R' H OH R (ii) Draw the structure of the organic product formed when the following are reacted: C O H and MgBr C OH H (iii) Draw the displayed formula of the chloroalkane which can be used to form a Grignard reagent that would react with propanone to give 2-methylpentan-2-ol. C H H C H H C H H H Cl [4] Dry ether Grignard reagent

(b) The following shows a series of reactions that compound M can undergo. C O-Na+O CH2 CH2 C OH CH3 C I H CH2 C CH3 H Cl CH2 CH CH3 OH CH2 C CH3 C H CH2 C CH3 H Cl Compound M LiAlH4 in dry ether P Q C17H27ON N O + KMnO4, H2SO4 (aq), heat Br2, Fe, Heat Step 1 Step 2

(i) Draw the structures of compounds N to Q in the boxes provided. N/O: COOH HOOC N/O: P: Q: CH2 CH2Br CH OH CH3 C CH3 Br C Br H CH2 CH CH3 Cl (ii) State the reagents and conditions for step 1 and 2: Step 1: Aqu