19.1. Primary amines
A subsection of Chemistry, 9701, through 19. Nitrogen compounds
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Ethylamine and phenylamine are primary amines. NH2 phenylamine ethylamine CH3CH2NH2 These two compounds are synthesised by different methods. Several methods can be used to form ethylamine. Ethylamine forms when ethanamide, CH3CONH2, is reduced by LiAl H4. Write an equation for this reaction. Use to represent one atom of hydrogen from the reducing agent. Ethylamine is a product of the reaction of bromoethane with ammonia. Name the mechanism of this reaction and state the conditions used. mechanism conditions The reaction in also forms secondary and tertiary amines. Suggest the identity of a secondary or tertiary amine formed by the reaction in . Ethylamine is a weak base. State the relative basicities of ammonia, ethylamine and phenylamine. Explain your answer. least basic most basic Pure phenylamine, C6H5NH2, can be prepared from benzene in two steps. Draw the structure of the intermediate compound. Suggest reagents and conditions for each step. shows some reactions of phenylamine. NH2 N2Cl OH N N phenylamine reaction 1 reaction 2 excess Br2room temperature reaction 4 W reaction 3 X Y benzenediazonium chloride Draw the structure of W, the organic product of reaction 1. State the reagents used in reaction 2. Benzenediazonium chloride, C6H5N2Cl, and X react together in reaction 4 to form Y, an azo compound. Name X, the organic product of reaction 3. State the necessary conditions for reaction 4 to occur. Suggest a use for Y. Methylamine, CH3NH2, is another primary amine. CH3NH2 can act as a monodentate ligand. Define monodentate ligand. Cu2+reacts with CH3NH2 to form [Cu(CH3NH2)2(H2O)4]2+. Draw three-dimensional diagrams to show the two geometrical isomers of [Cu(CH3NH2)2(H2O)4]2+. Cu Cu State the coordination number of copper in [Cu(CH3NH2)2(H2O)4]2+. Cd2+ions form tetrahedral complexes with CH3NH2, OH– and Cl – ions, as shown in equilibria 1, 2 and 3. equilibrium 1 Cd2++ 4CH3NH2[Cd(CH3NH2)4]2+Kstab = 3.3 × 106 equilibrium 2 Cd2++ 4OH–Cd(OH)4 2–Kstab = 5.0 × 108 equilibrium 3 Cd2++ 4Cl –CdCl 4 2–Kstab = 6.3 × 102 Give the units of Kstab for equilibrium 1. Write an expression for Kstab for equilibrium 3. Kstab = A solution of Cl –is added to Cd2+and allowed to reach equilibrium. The equilibrium concentrations are given. [Cd2+] = 0.043 mol dm–3 [Cl –] = 0.072 mol dm–3 Use your expression in to calculate the concentration of CdCl 4 2–in the equilibrium mixture. [CdCl 4 2–] = mol dm–3 When CH3NH2is added to Cd2+, a mixture of [Cd(CH3NH2)4]2+and [Cd(OH)4]2–forms. Suggest how the [Cd(OH)4]2–is formed. Cd2+exists as a complex ion, [Cd(H2O)6]2+. Identify the most stable and the least stable of the complexes in Table 4.1 by placing one tick (3) in each column. Explain your answer. Table 4.1 complex most stable least stable [Cd(H2O)6]2+[Cd(OH)4]2–[Cd(CH3NH2)4]2+2–explanation
9701_m23_qp_42
THEORY
2023
Paper 4, Variant 2
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