14.2. Alkenes
A subsection of Chemistry, 9701, through 14. Hydrocarbons
Listing 10 of 191 questions
The structural and displayed formulae of three aromatic compounds, A, B and C, are shown in Fig.7.1. C6H5CH2CH2COOH A HO O OH CH3C6H4OH B C6H5CH2CHCl CO2H C HO O Cl Compare the relative acidities of A, B and C. > > most acidic least acidic Explain your answer. Methylbenzene, C6H5CH3, can be made from benzene by an electrophilic substitution reaction. Identify a compound that reacts with benzene to form methylbenzene. Identify the catalyst used. compound catalyst  The first step in the reaction is the generation of the CH3 + electrophile. Write an equation for the reaction that generates this electrophile. Describe the mechanism for the reaction between benzene and the CH3 + electrophile. Include all relevant curly arrows and charges. CH3 +  Identify a suitable reagent to oxidise methylbenzene to benzoic acid, C6H5COOH. Write an equation for this reaction using to represent one atom of oxygen from the oxidising agent. reagent equation  Methylbenzene and benzoicacid are both nitrated with a mixture of concentrated nitricacid and sulfuricacid to give mononitrated products. The structural formulae of these products are CH3C6H4NO2 and HOOCC6H4NO2 respectively. Draw the structures of these two products.  A reaction scheme is shown in Fig.7.2. CH3C6H4NO2 CH3C6H4NH2 reaction 1 reaction 2 CH3C6H4OH CH3C6H4N2 +Cl – dye molecule Y reaction 3 reaction 4 Describe the reagents and conditions to produce CH3C6H4N2 +Cl – from CH3C6H4NH2 in reaction2. reagents conditions  Describe how CH3C6H4OH can be produced from CH3C6H4N2 +Cl – in reaction3. Draw the structure of the dye moleculeY formed when CH3C6H4N2 +Cl – and CH3C6H4OH react together in reaction4. Describe the conditions for this reaction. structure conditions  
9701_w22_qp_42
THEORY
2022
Paper 4, Variant 2
View
Compounds C and D are alkenes with the same molecular formula, C5H10. C D Give the systematic name of D. Explain why C and D do not show geometrical (cis/trans) isomerism. Draw the structure of a molecule that is a positional isomer of C and D. Give the structural formula of the compound formed when D reacts with H2in the presence of a Pt catalyst. C can form an addition polymer. Draw the structure of one repeat unit of this addition polymer. The mass spectrum of C shows a molecular ion peak at m/e = 70. This peak has a relative intensity of 48.7. The relative intensity of the [M+1] peak is 2.7. Show that this information is consistent with the molecular formula of C. C and D both react with HBr. C reacts with HBr to form E. Complete the diagram in to show the mechanism for this reaction. Draw the structure of the organic intermediate. Include charges, dipoles, lone pairs of electrons and curly arrows, as appropriate. C H – Br E Br D reacts with HBr to produce F, a chiral bromoalkane. Draw the structure of F. F Explain why the reaction of HBr with C and D produces different major products. C D C can be used to form H. H O O One possible synthesis of H is shown in . Different portions of C are used in reactions 1 and 3. Some of the products are then combined to produce H. does not show any of the inorganic products of the reactions. C E G + J + G then: J O Br HBrreaction 1 reaction 2 reaction 3 reaction 4 H O O
9701_w23_qp_21
THEORY
2023
Paper 2, Variant 1
View
Compounds C and D are alkenes with the same molecular formula, C5H10. C D Give the systematic name of D. Explain why C and D do not show geometrical (cis/trans) isomerism. Draw the structure of a molecule that is a positional isomer of C and D. Give the structural formula of the compound formed when D reacts with H2in the presence of a Pt catalyst. C can form an addition polymer. Draw the structure of one repeat unit of this addition polymer. The mass spectrum of C shows a molecular ion peak at m/e = 70. This peak has a relative intensity of 48.7. The relative intensity of the [M+1] peak is 2.7. Show that this information is consistent with the molecular formula of C. C and D both react with HBr. C reacts with HBr to form E. Complete the diagram in to show the mechanism for this reaction. Draw the structure of the organic intermediate. Include charges, dipoles, lone pairs of electrons and curly arrows, as appropriate. C H – Br E Br D reacts with HBr to produce F, a chiral bromoalkane. Draw the structure of F. F Explain why the reaction of HBr with C and D produces different major products. C D C can be used to form H. H O O One possible synthesis of H is shown in . Different portions of C are used in reactions 1 and 3. Some of the products are then combined to produce H. does not show any of the inorganic products of the reactions. C E G + J + G then: J O Br HBrreaction 1 reaction 2 reaction 3 reaction 4 H O O
9701_w23_qp_23
THEORY
2023
Paper 2, Variant 3
View
shows two reactions of ethanedioic acid, HOOCCOOH. an excess of SOCl 2 G 2CO2 + H2O HOOCCOOH Draw the organic product G in the box in . In , SOCl 2 is given as the reagent that reacts with HOOCCOOH to produce G. Identify a different reagent that also reacts with HOOCCOOH to produce G. Identify two different reagents that oxidise HOOCCOOH to form carbon dioxide and water. HOOCCOOH ionises as shown. HOOCCOOH HOOCCOO– + H+ HOOCCOOH is a much stronger acid than methanoic acid, HCOOH. Suggest an explanation for this difference in acidity. Benzene-1,4-dicarboxylic acid, HOOCC6H4COOH, can be made from benzene, C6H6, in two steps as shown in . J step 1 step 2 HOOC COOH benzene-1,4-dicarboxylic acid Suggest the identity of J by drawing its structure in the box in . Identify the reagents and conditions for step 1 and step 2. step 1 step 2 Draw the structure of exactly one repeat unit of the polymer formed when benzene-1,4-dicarboxylic acid reacts with ethane-1,2-diol, HOCH2CH2OH. The linkage formed between the monomers should be shown fully displayed. State the type of polymerisation that occurs when benzene-1,4-dicarboxylic acid reacts with ethane-1,2-diol and name the linkage formed between the monomers. type of polymerisation linkage
9701_w24_qp_42
THEORY
2024
Paper 4, Variant 2
View
Questions Discovered
191