The reducing agent LiAlH4 can be synthesised by reacting aluminium chloride with lithium hydride, LiH.

1. (a)(i)At 200 °C, aluminium chloride exists as Al2Cl6(g). Draw the structure of Al₂Cl6(g), showing fully any coordinate (dative covalent) bonds in the molecule.[2]
2. (a)(ii)At 1000°C, aluminium chloride exists as AlCl3(g). State the bond angle in AlCl3(g).[1]
3. (a)(iii)Lithium hydride contains the ions Li⁺ and H⁻. State the electronic configuration of these two ions. Li⁺ ............................ H⁻ ............................[1]
4. (a)(iv)LiAlH₄ decomposes slowly to form LiAl(s) and H2(g). LiAlH₄(s) → LiAl(s) + 2H2(g). LiAl(s) shows metallic bonding. Describe metallic bonding.[1]
5. (b)LiAlH4 cannot be used in aqueous solution because it reacts with water to produce LiOH(aq), H2(g) and a white precipitate which is soluble in excess sodium hydroxide. Identify the white precipitate.[1]
6. (c)(i)Two students try to prepare 2-hydroxybutanoic acid in the laboratory. [Figure 3.1: Diagram of 2-hydroxybutanoic acid] Both students oxidise butane-1,2-diol to form P in reaction 1. One student then reduces P using LiAlH4. Q is formed. The other student reduces P using NaBH4. R is formed. [Figure 3.2: Reaction scheme showing butane-1,2-diol to P, then P to Q with LiAlH4 and P to R with NaBH4] State the reagents and conditions required for reaction 1.[2]
7. (c)(ii)Only one of the students successfully prepares 2-hydroxybutanoic acid. Identify which of Q or R is 2-hydroxybutanoic acid and explain the difference between reactions 2 and 3.[2]
8. (c)(iii)A third student prepares 2-hydroxybutanoic acid using propanal as the starting material. In step 1 the student reacts propanal with a mixture of NaCN and HCN. [Figure 3.3: Reaction scheme from propanal to S, then S to 2-hydroxybutanoic acid] Draw the mechanism for the reaction of propanal with the mixture of NaCN and HCN to form S. • Identify the ion that reacts with propanal. • Draw the structure of the intermediate of the reaction. • Include all charges, partial charges, lone pairs and curly arrows.[4]
9. (c)(iv)Complete the equation for the reaction in step 2, when S is heated under reflux with HCl(aq). C2H5CH(OH)CN + ................................... → C2H5CH(OH)COOH + ...................................[1]
10. (c)(v)The infrared spectrum of an organic compound is shown. The organic compound is either S or 2-hydroxybutanoic acid. [Figure 3.4: Infrared spectrum graph] Deduce the identity of the compound. Give two reasons for your answer. In your answer, identify any relevant absorptions above 1500 cm⁻¹ in the spectrum and the bonds that correspond to these absorptions.[2]

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