2021 YIJC Prelim P3 Answer

2021 YIJC Preliminary Examination H2 Chemistry Paper 3 with Mark Schemes Section A Answer all the questions in this section. 1 5-bromo-4-methyl-3-penten-2-ol can be made from 5 -bromo-4-methyl-3-penten-2-one in one step. The molecules are labelled as A and B respectively in Fig. 1.1. Fig. 1.1 (a) (i) State the total number of σ and π bonds in a molecule of A. [1] 16 σ and 2 π (ii) Describe the hybridisation of the orbitals and the bonds between the carbon atoms of the C=C double bond in A. [2] sp2 1 σ bond – head on overlap of sp2 hybridised orbitals of the 2 C atoms. 1 π bond – side on overlap of the unhybridised p orbitals of the 2 C atoms. (ii) State the total possible number of stereoisomers that can be exhibited by B. [1] 1 chiral carbon + 1 C=C that can exhibit cis trans isomerism = 22 = 4 (iv) Sodium borohydride is used to reduce A to form B.

Explain why only the ketone functional group in A undergoes reduction with sodium borohydride but the alkene functional group does not. [2] The carbonyl carbon in ketone is electron deficient due to the difference in electronegativity between C and O. Hence it attracts the :H– nucleophile. The C=C in alkenes are electron rich and will repel the negatively charged :H– nucleophile. (v) Describe a chemical test that can be used to distinguish A and B. [2] Test : 2,4 – DNPH, rtp A: orange ppt formed B : no orange ppt formed Test : KMnO4 , H2SO4(aq), heat A: purple KMnO4 remains B : purple KMnO4 decolourises Test : K2Cr2O7 , H2SO4(aq), heat A: orange K2Cr2O7 remains B : orange K2Cr2O7 turns green Test : Na, rtp A: no effervescence B : effervescence observed, gas extinguish lighted splint with a pop sound Test : anhydrous PCl5 , SOCl2, rtp A: no white fumes B : white fumes observed

Compound E can be made from A by a four-step synthesis. Fig. 1.2 (b) (i) Suggest the reagents and conditions for steps 2 and 3 in Fig. 1.2. [2] Step 2: ethanolic KCN, heat Step 3: H2SO4(aq), heat (ii) Compound E is neutral and does not react with 2,4 -dinitrophenylhydrazine or sodium metal. Draw the structures for compounds C, D and E. [3] C: D:

E: (iii) 5-chloro-4-methyl-3-penten-2-ol can be used as the reactant in step 2 instead of B in the reaction scheme in Fig. 1.2. Suggest how the rate of reaction will change if 5-chloro-4-methyl-3-penten-2-ol is used. [2] The chlorine atom has a smaller valence orbital / less diffused/ smaller size / smaller radius/ smaller electron cloud size than that of bromine. Hence, the effectiveness of orbital overlap is more effective between C and Cl, making the C–Cl bond stronger . Hence more energy is required to break the C–Cl bond and the reaction will be slower. The Canniz zaro reaction is a base catalysed reaction, which involves the disproportionation of aldehydes to form a carboxylic acid and an alcohol. 2RCHO  RCOOH + RCH2OH The mechanism of the Cannizzaro reaction involves 4 steps. A nucleophilic attack on the carbonyl carbon of the aldehyde produce s a dianion via a 2-step process. In step 3, the dianion reacts with another aldehyde molecule to form the carboxylate and alkoxide ions.

In step 4, both the carboxylate and alkoxide ions are protonated to form the products. (c) (i) Define the term disproportionation, with reference to the Cannizzaro reaction. [2] Disproportionation is a redox reaction where one species is oxidised and reduced simultaneously. With reference to the Cannizzaro reaction, the aldehyde molecules undergo oxidation to form carboxylic acid and undergo reduction to form alcohol. OR The aldehyde molecules undergo oxidation to form carboxylic acid and undergo reduction to form alcohol simultaneously. (ii) In step 1 of the mechanism, the OH– ion acts as a nucleophile and attacks the carbonyl carbon to fo rm an intermediate. In step 2 , the second OH – ion abstracts a proton from the hydroxyl group on the intermediate to form the dianion. Showing any relevant lone pairs, dipoles and charges, indicate the movement of electrons with curly arrows to outline the mechanism involved in the first 2 steps of the Cannizzaro reaction. [2]

(iii) Deduce the products of the reactio n when benzaldehyde undergoes the Cannizzaro reaction. [2] [Total: 21]

2 (a) This part of the question is about compounds of Group 17 elements and period 3 elements. (i) Using data from the Data Booklet, state and explain how the thermal stability of the Group 17 hydrides vary down the group. [3] 2HX(g)  H2(g) + X2(g) (X = Cl, Br, I) Down the group, size of valence orbital of X increases. Overlap of orbitals between H and X becomes less effective. This leads to the H−X bond becoming weaker , as shown by the decreasing bond energy values. Quote bond energy values Bond Bond Energy / kJ mol-1 H–Cl 431 H–Br 366 H–I 299 Less energy is required to break the covalent bond H−X for thermal decomposition. Hence, thermal stability decreases. (ii) Describe the reactions, if any, of A lCl3 and PCl5 with water, stating the pH of the resulting solutions. Write equations where appropriate. [4] AlCl3 hydrolyses partially/slightly in water to form a slightly acidic solution of pH 3. Hydration: AlCl3(aq) + 6H2O(l)  [Al(H2O)6]3+(aq) + 3Cl-(aq) Hydrolysis: [Al(H2O)6]3+(aq) ⇌ [Al(H2O)5(OH)]2+(aq) + H+(aq) PCl5 hydrolyses readily/completely in water to form a strongly acidic solution of pH 1-2. PCl5 + 4H2O  H3PO4 + 5HCl

F, G and H are oxides of period 3 elements. F and H are highly soluble in water whereas G is insoluble in water. When treated with H2SO4(aq), only F and G reacts to form a colourless solution. When treated with NaOH(aq), only G and H reacts to form colourless solution J and K respectively. K forms white precipitate when treated with acidified aqueous barium nitrate. (iii) Suggest the identity of the oxides F, G and H. [2] F: Na2O G: Al2O3 H: SO3 (accept SO2) (iv) Hence, write equations for the reactions of G and H with NaOH(aq). [2] For G: Al2O3 + 2NaOH + 3H2O  2Na[Al(OH)4] (accept ionic eqn: Al2O3 + 2OH- + 3H2O  2[Al(OH)4]-) For H: SO3 + 2NaOH  Na2SO4 + H2O (accept SO2 + 2NaOH  Na2SO3 + H2O if student used SO2)

(b) Compounds of period 3 elements , such as sulfuric acid, are often used in organic reactions. For example, sulfonation is a reaction to incorporate sulfonic acid functional group (– SO3H) into a molecule. Both alkenes and arenes can undergo sulfonation under suitable reagents and conditions. (i) One example of sulfonation of alkene is the reaction of concentrated sulfuric acid, H2SO4 with alkenes via an addition reaction. With propene, isomer L is produced rather than isomer M. Fig. 2.1 By considering the mechanism and intermediates of the reaction, explain the preferential production of isomer L. [3] Carbocation that form L Carbocation that form M In the electrophilic addition mechanism, t he carbocation forming F is a secondary carbocation which has 1 more electron -donating alkyl group whereas the carbocation forming G is a primary carbocation . Hence, the positive charge on the secondary carbocation is dispersed more effectively, and it is more stable.

In sulfonation of benz