2022 DHS Y6 H2 Prelim Paper 3 Suggested Solutions

© DHS 2022 9729/03 [Turn over Suggested Solutions DUNMAN HIGH SCHOOL Preliminary Examination Year 6 H2 CHEMISTRY Paper 3 Free Response Questions Candidates answer on the Question Paper. Additional Materials: Data Booklet 9729/03 21 September 2022 2 hours READ THESE INSTRUCTIONS FIRST Write your centre number, index number, name and class at the top of this page. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all questions in the spaces provided on the Question Paper. If additional space is required, you should use the pages at the end of this booklet. The question number must be clearly shown. Section A Answer all questions. Section B Answer one question. A Data Booklet is provided. The use of an approved scientific calculator is expected, where appropriate. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use Section A 1 20 2 20 3 20 Section B 4 / 5 20 Total 80 This document consists of 23 printed pages.

2 © DHS 2022 9729/03 Section A Answer all the questions from this section. 1 (a) Alkynes is a class of organic compounds with the general formula, CnH2n−2. (i) Describe what is meant by sp hybridisation with reference to one carbon atom in ethyne, C2H2. Draw the hybrid orbitals of the carbon atom. [2] One 2s and one 2p atomic orbitals are mixed to form two 2sp hybrid orbitals that are degenerate / of equal energy. internuclear axis (ii) Use relevant radius values from the Data Booklet to calculate the bond length of a single carbon-hydrogen bond. Show your working clearly. [1] From the Data Booklet, single covalent radius of hydrogen = 0.037 nm single covalent radius of carbon = 0.077 nm Bond length = sum of covalent radii = 0.037 nm + 0.077 nm = 0.114 nm (iii) Table 1.1 shows the carbon-hydrogen bond length in ethene and ethyne. Table 1.1 molecule carbon-hydrogen bond length/ nm ethene 0.109 ethyne 0.106 With reference to Table 1.1, state which carbon -hydrogen bond is stronger. Use the concept of hybridisation to explain the difference in bond length of the carbon-hydrogen bond between these two molecules. [2] The carbon-hydrogen bond of ethyne is stronger. The sp hybridised carbon atom in ethyne has a higher percentage s character than the sp 2 hybridised carbon atom in ethene. Hence, the extent of orbital overlap between the sp hybridised carbon atom and H atom is greater, resulting in a shorter bond length. (iv) Write a balanced equation for the complete combustion of propyne, C3H4. [1] C3H4 + 4O2 → 3CO2 + 2H2O (v) A sample of propyne was burned in excess oxygen. When the remaining gases were passed through aqueous sodium hydroxide, the gas volume was reduced by 0.450 dm3. Calculate the mass of propyne in the sample.

3 © DHS 2022 9729/03 [Turn over Assume all gas volumes were measured at r.t.p. [2] Since carbon dioxide is the only acidic gas, volume of CO2 = 0.450 dm3 3CO2 ≡ C3H4 By volume ratio, volume of C3H4 = 0.150 dm3 Moles of C3H4 = 0.15 24 = 0.00625 mol Mass of C3H4 = 0.00625 ((4 1.0) (3 12.0))  +  = 0.250 g (b) The use of hydrogen gas with Lindlar’s catalyst is a selective method which reduces alkynes to form the cis-isomer of alkenes. As an example, but-2-yne, C4H6, can be reduced to give cis-but-2-ene only. H H H2(g) Lindlar's catalyst Suggest the structure of the alkene formed when compound A is reduced with the use of Lindlar’s catalyst. A [1] (c) Compound B has the following structure. 1 2 B (i) State the isomeric relationship between B and your answer in (b). [1] Chain / constitutional isomers

4 © DHS 2022 9729/03 (ii) A student made the following deductions about compound B: “Since carbon atoms labelled 1 and 2 are chiral and there are two carbon-carbon double bonds, compound B has 16 possible stereoisomers.” Explain where the student has gone wrong in his deductions. [3] Carbon 1 is not chiral as it is bonded to two identical methyl groups. For the C=C bond in the ring, only the cis-isomer exists. In the presence of a small six-membered ring, the trans-isomer will be highly strained and unstable. With one chiral centre and one C=C bond not located in a ring, there is a total of 22 = 4 possible stereoisomers for compound B. (iii) Draw all the organic products that are formed when compound B is heated with acidified potassium manganate(VII). [1] O OH O OH O OH O + Marker’s Comments Typical mistakes include: • Oxidation of the C=C bond to aldehyde functional group instead of carboxylic acid functional group • Missing ethanoic acid as one of the organic products • Missing methyl groups or carbon atoms • Five bonds around carbon in C=O of the ketone functional group (d) Compound C is an isomer of compound B. C Draw the structure of the major product formed when compound C is reacted with excess HBr(g). Explain your answer. [2]

5 © DHS 2022 9729/03 [Turn over Br Br The tertiary carbocation is more stable than the secondary carbocation formed as there are more electron –donating alkyl groups to disperse the positive charge on each carbon atom. (e) Lithium aluminium hydride, LiA lH4, is a reducing agent commonly used in organic chemistry. (i) Assuming LiAlH4 as a source of hydride (H−) ions, suggest why the reduction of alkynes using LiAlH4 is likely not a suitable method. [1] The π electrons / triple bond / C≡C bond in alkynes will repel the negatively charged H−. (ii) LiAlH4 can be synthesised from aluminium chloride, which exists as a dimer, Al2Cl6, at room temperature. Draw a dot-and-cross diagram to illustrate the bonding present in Al2Cl6. [1] Al Al Cl Cl Cl Cl Cl Cl xx xx xx xx xx xx xx xx xxxx xx xx xx xx xxxxxx x x xxx x x x (iii) When heated, LiAlH4 decomposes to LiAl(s) as one of its products. LiAl(s) has a melting point of 718 C. It can conduct electricity when in solid and molten states. Suggest the structure of LiAl(s) and describe the bonding present. [2] Giant metallic structure. Strong electrostatic forces of attraction between lattice of Li + and A l3+/ positively-charged ions and the sea of delocalised electrons. [Total: 20]

6 © DHS 2022 9729/03 2 (a) 1,2–diols are common precursors used in many pharmaceuticals, agrochemicals, and natural products. Fig. 2.1 shows a pinacol coupling reaction which involves the homo–coupling* of a carbonyl compound to produce a symmetrically substituted 1,2–diol. *Two identical molecules react to form a different one. 2 R O + Mg R O O R R R OH OH + Mg2+ radical ion intermediates H2O −Mg(OH)2 1,2 diol Fig. 2.1 The first step is single electron transfer involving the carbonyl group, which generates radical ion intermediates that couple via carbon –carbon bond formation to give a 1,2–diol. (i) State the role of magnesium in the first step of Fig . 2.1 and give a reason for its suitability in this reaction. [2] Magnesium is functioning as a reducing agent in the first step. It is suitable due to its relatively low ionisation energies/ low electronegativity. (ii) It is possible to synthesise a desired unsymmetrical diol using methods similar to the pinacol coupling reaction but a mixture of diols will be obtained. Explain why a mixture of diols is formed. Suggest why this is unfavourabl