14. Hydrocarbons
A section of Chemistry, 9701
Listing 10 of 313 questions
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
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Capsaicin is found in chilli peppers. CH3O HO O N H capsaicin You should assume the CH3O group is unreactive in the reactions involved in this question. Name all the functional groups in capsaicin in addition to the CH3O group. Complete the equation for the reaction of capsaicin with an excess of Br2in the dark. Draw the structure of the organic product in the labelled box. CH3O HO O capsaicin organic product + ………………… + ………Br2 N H Capsaicin is heated with an excess of hydrogen gas in the presence of platinum metal. The six-membered ring reacts in the same way as benzene under these conditions. Draw the structure of the organic product formed. When capsaicin is treated with reagent J under suitable conditions one of the products is methylpropanoic acid, CH3CH(CH3)COOH. Identify reagent J and any necessary conditions. There are three different peaks in the proton (1H) NMR spectrum of CH3CH(CH3)COOH in CDCl 3. Table 8.1 environment of proton example chemical shift range δ / ppm alkane –CH3, –CH2–, >CH– 0.9–1.7 alkyl next to C=O CH3–C=O, –CH2–C=O, >CH–C=O 2.2–3.0 alkyl next to aromatic ring CH3–Ar, –CH2–Ar, >CH–Ar 2.3–3.0 alkyl next to electronegative atom CH3–O, –CH2–O, –CH2–Cl 3.2–4.0 attached to alkene =CHR 4.5–6.0 attached to aromatic ring H–Ar 6.0–9.0 aldehyde HCOR 9.3–10.5 alcohol ROH 0.5–6.0 phenol Ar–OH 4.5–7.0 carboxylic acid RCOOH 9.0–13.0 Use Table 8.1 to complete Table 8.2 and state: ● the typical proton (1H) chemical shift values (δ) for the protons ● the splitting pattern (singlet, doublet, triplet, quartet or multiplet) shown by each peak ● the explanation for the splitting patterns of the CH3 protons and the CH proton. Table 8.2 environment δ / ppm splitting pattern explanation for splitting pattern CH3 CH COOH Capsaicin is heated with an excess of hot aqueous NaOH. CH3O HO O N H capsaicin Draw the structures of the two organic products H and K. H C8H10NO2Na K C10H17O2Na Name the two types of reaction occurring in . Draw the structure of the organic product L formed when capsaicin is treated with LiAl H4 in dry ether. L
9701_w23_qp_41
THEORY
2023
Paper 4, Variant 1
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Capsaicin is found in chilli peppers. CH3O HO O N H capsaicin You should assume the CH3O group is unreactive in the reactions involved in this question. Name all the functional groups in capsaicin in addition to the CH3O group. Complete the equation for the reaction of capsaicin with an excess of Br2in the dark. Draw the structure of the organic product in the labelled box. CH3O HO O capsaicin organic product + ………………… + ………Br2 N H Capsaicin is heated with an excess of hydrogen gas in the presence of platinum metal. The six-membered ring reacts in the same way as benzene under these conditions. Draw the structure of the organic product formed. When capsaicin is treated with reagent J under suitable conditions one of the products is methylpropanoic acid, CH3CH(CH3)COOH. Identify reagent J and any necessary conditions. There are three different peaks in the proton (1H) NMR spectrum of CH3CH(CH3)COOH in CDCl 3. Table 8.1 environment of proton example chemical shift range δ / ppm alkane –CH3, –CH2–, >CH– 0.9–1.7 alkyl next to C=O CH3–C=O, –CH2–C=O, >CH–C=O 2.2–3.0 alkyl next to aromatic ring CH3–Ar, –CH2–Ar, >CH–Ar 2.3–3.0 alkyl next to electronegative atom CH3–O, –CH2–O, –CH2–Cl 3.2–4.0 attached to alkene =CHR 4.5–6.0 attached to aromatic ring H–Ar 6.0–9.0 aldehyde HCOR 9.3–10.5 alcohol ROH 0.5–6.0 phenol Ar–OH 4.5–7.0 carboxylic acid RCOOH 9.0–13.0 Use Table 8.1 to complete Table 8.2 and state: ● the typical proton (1H) chemical shift values (δ) for the protons ● the splitting pattern (singlet, doublet, triplet, quartet or multiplet) shown by each peak ● the explanation for the splitting patterns of the CH3 protons and the CH proton. Table 8.2 environment δ / ppm splitting pattern explanation for splitting pattern CH3 CH COOH Capsaicin is heated with an excess of hot aqueous NaOH. CH3O HO O N H capsaicin Draw the structures of the two organic products H and K. H C8H10NO2Na K C10H17O2Na Name the two types of reaction occurring in . Draw the structure of the organic product L formed when capsaicin is treated with LiAl H4 in dry ether. L
9701_w23_qp_43
THEORY
2023
Paper 4, Variant 3
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Benzene, C6H6, reacts with chloroethane, C2H5Cl, in the presence of a suitable catalyst to form ethylbenzene, C6H5C2H5. In the presence of the catalyst, the ion C2H5 + is formed. This ion reacts with benzene. Complete the equation for the reaction of C2H5Cl with this catalyst to form C2H5 + as one product. C2H5Cl + C2H5 + + Ethylbenzene reacts with more C2H5Cl, forming a mixture containing 1,2-diethylbenzene and 1,4-diethylbenzene. Draw the structures of 1,2-diethylbenzene and 1,4-diethylbenzene. 1,2-diethylbenzene 1,4-diethylbenzene Explain why there is very little 1,3-diethylbenzene in the product mixture. 1,2-diethylbenzene can be oxidised to benzene-1,2-dioic acid, C6H4(COOH)2. COOH COOH benzene-1,2-dioic acid State the reagent and conditions used for this reaction. Complete the overall equation for this reaction. An atom of oxygen from the oxidising agent is represented as . All of the atoms in the two ethyl groups are fully oxidised in this reaction. + C6H4(COOH)2 + + (1,2-diethylbenzene) Predict the number of peaks in the carbon-13 NMR spectrum of benzene-1,2-dioic acid. The proton (1H) NMR spectra of ethylbenzene, C6H5C2H5, in CDCl 3 and of benzene-1,2-dioic acid, C6H4(COOH)2, in CDCl 3 are shown. They have not been identified. δ / ppm δ / ppm Explain the use of CDCl 3, instead of CHCl 3, as the solvent when obtaining these spectra. Identify the substance shown by the spectrum in , and complete Table 6.1. substance Table 6.1 peak at δ = 1.2 peak at δ = 2.6 name of splitting pattern group responsible for peak explanation of splitting pattern Identify the substance shown by the spectrum in , and complete Table 6.2. substance Table 6.2 peak at δ = 7.8 peak at δ = 13.1 group responsible for peak When D2O is used as a solvent, the spectrum obtained is different from the spectrum in . Describe this difference and explain your answer. Benzene-1,2-dioic acid can be used to produce K. COOH COOH C C O O K O benzene-1,2-dioic acid heat Suggest the name of this type of reaction.
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THEORY
2024
Paper 4, Variant 1
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Benzene, C6H6, reacts with chloroethane, C2H5Cl, in the presence of a suitable catalyst to form ethylbenzene, C6H5C2H5. In the presence of the catalyst, the ion C2H5 + is formed. This ion reacts with benzene. Complete the equation for the reaction of C2H5Cl with this catalyst to form C2H5 + as one product. C2H5Cl + C2H5 + + Ethylbenzene reacts with more C2H5Cl, forming a mixture containing 1,2-diethylbenzene and 1,4-diethylbenzene. Draw the structures of 1,2-diethylbenzene and 1,4-diethylbenzene. 1,2-diethylbenzene 1,4-diethylbenzene Explain why there is very little 1,3-diethylbenzene in the product mixture. 1,2-diethylbenzene can be oxidised to benzene-1,2-dioic acid, C6H4(COOH)2. COOH COOH benzene-1,2-dioic acid State the reagent and conditions used for this reaction. Complete the overall equation for this reaction. An atom of oxygen from the oxidising agent is represented as . All of the atoms in the two ethyl groups are fully oxidised in this reaction. + C6H4(COOH)2 + + (1,2-diethylbenzene) Predict the number of peaks in the carbon-13 NMR spectrum of benzene-1,2-dioic acid. The proton (1H) NMR spectra of ethylbenzene, C6H5C2H5, in CDCl 3 and of benzene-1,2-dioic acid, C6H4(COOH)2, in CDCl 3 are shown. They have not been identified. δ / ppm δ / ppm Explain the use of CDCl 3, instead of CHCl 3, as the solvent when obtaining these spectra. Identify the substance shown by the spectrum in , and complete Table 6.1. substance Table 6.1 peak at δ = 1.2 peak at δ = 2.6 name of splitting pattern group responsible for peak explanation of splitting pattern Identify the substance shown by the spectrum in , and complete Table 6.2. substance Table 6.2 peak at δ = 7.8 peak at δ = 13.1 group responsible for peak When D2O is used as a solvent, the spectrum obtained is different from the spectrum in . Describe this difference and explain your answer. Benzene-1,2-dioic acid can be used to produce K. COOH COOH C C O O K O benzene-1,2-dioic acid heat Suggest the name of this type of reaction.
9701_w24_qp_43
THEORY
2024
Paper 4, Variant 3
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Cyclohexane is a colourless liquid used in industry to produce synthetic fibres. A reaction scheme involving cyclohexane is shown. reaction 1 reaction 2 Cl Reaction1 involves a free radical substitution mechanism. State the essential condition required for reaction1 to occur. Complete the table to give details of the mechanism in reaction1. name of step reaction Cl 2 2Cl • propagation + Cl • + • Cl + Cl 2 + Cl • • termination + Cl • •  Name the type of reaction that occurs in reaction2. The product of reaction2 is cyclohexene. Cyclohexene can be converted into adipicacid (hexanedioicacid), HO2C(CH2)4CO2H. Identify the reagents and conditions for the conversion of cyclohexene into adipicacid. Suggest three main differences between the infra-red spectra of cyclohexene and adipicacid. In each case, identify the bond responsible and its characteristic absorption range (in wavenumbers).  
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THEORY
2018
Paper 2, Variant 2
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