RI 2024 Alkenes Answer Key

-1- Answers to Practice Questions 1(a)(i) Molecular formula of Q is C5H8. 1(a)(ii) Isomers of Q 1(b) Squalene has 6 π bonds per molecule. 1(c) cis-pent-2-ene > trans-pent-2-ene > 2-methylbut-1-ene 2(a) A: C C CH3 H CH3 H B: C C H CH3 CH3 H C: C C H H CH2CH3 H cis-but-2-ene trans-but-2-ene but-1-ene D: C C H H CH3 CH3 E: F: CH3 2–methylpropene cyclobutane methylcyclopropane 2(b) G: C C * H C H H Br C H H H H H Br 1,2-dibromobutane H: C C H H H CH H H H CH H H 2-methylpropane 2(c) A (i.e. cis-but-2-ene) is slightly polar. In addition to instantaneous dipole -induced dipole attractive forces present between the molecules, there are also permanent dipole -permanent dipole attractive forces. B (i.e. trans-but-2-ene) is non-polar. Only instantaneous dipole-induced dipole attractive forces exist between the molecules. Since more energy is needed to overcome the stronger intermolecular forces in A during boiling, the boiling point of A is higher than that of B. 2(d) Reagent Alkene (i) cold, alkaline KMnO4(aq) (ii) hot KMnO4(aq) made alkaline with NaOH(aq) (iii) hot KMnO4(aq) acidified with H2SO4(aq) A C C CH3 H CH3 H OHHO CH3CO2−Na+ CH3CO2H B C C H CH3 CH3 H OHHO CH3CO2−Na+ CH3CO2H C C C H H CH2CH3 H OHHO CH3CH2CO2−Na+ CH3CH2CO2H

-2- D C C H H CH3 CH3 OHHO CH3COCH3 CH3COCH3 3(a) Cl Br Br KOH in ethanol heat Br2 in CCl 4 3(b) OH BrBr KOH in ethanol heat Br2(aq) 3(c) OH OHexcess conc H2SO4 heat 1. cold conc H2SO4 2. H2O, heat 3(d) OH O OHexcess conc H2SO4 heat KMnO4(aq) H2SO4(aq) heat 4(a) Bubble a sample of each gas through Br2(aq) in separate test-tubes at room temperature. For ethene, there is (rapid) decolourisation of the orange Br2(aq). For ethane, there is no decolourisation of the orange Br2(aq). Equation: CH2=CH2 + Br2 + H2O → HOCH2CH2Br + HBr OR Add two drops of Br 2 (in CCl4) to each sample of gas already collected in a stoppered test -tube at room temperature and in the dark. Shake the re-stoppered test-tube. For ethene, there is (rapid) decolourisation of the orange-red Br2. For ethane, there is no decolourisation of the orange-red Br2. Equation: CH2=CH2 + Br2 → BrCH2CH2Br Other possible chemical tests: • KMnO4(aq) acidified with H2SO4(aq), heat (Note: DO NOT heat under reflux) • KMnO4(aq) and NaOH(aq), cold 4(b) Add two drops of KMnO 4 acidified with dilute sulfuric acid to each sample in a test -tube, and heat each reaction mixture in a hot water bath. (Note: DO NOT heat under reflux) For pent-2-ene, there is decolourisation of purple KMnO4. For pent-1-ene, there is decolourisation of purple KMnO 4 and evolution of a colourless gas (CO 2) which gives a white precipitate with limewater / Ca(OH)2(aq). Equations: + 4[O] KMnO4(aq), H2SO4(aq) heat O OH O OH +

-3- + 5[O] KMnO4(aq), H2SO4(aq) heat + O OH + CO2 H2O 4(c) Add two drops of KMnO 4 acidified with dilute sulfuric acid to each sample in a test -tube, and heat each reaction mixture in a hot water bath. (Note: DO NOT heat under reflux) For cyclohexa-1,4-diene, there is decolourisation of purple KMnO4. For cyclohexa-1,3-diene, there is decolourisation of purple KMnO4 and evolution of a colourless gas (CO2) which gives a white precipitate with limewater / Ca(OH)2(aq). Equations: + 8[O] KMnO4(aq), H2SO4(aq) heat 2 HO O OH O + 9[O] KMnO4(aq), H2SO4(aq) heat O O OH OH + + 2CO2 H2O 5(a) 5(b) CH3CH2CH=CHCH2CH3 + Br2 → CH3CH2CHBrCHBrCH2CH3 5(c) Electrophilic Addition Step 1: C C H CH3CH2 H CH2CH3 + Br C CH CH2CH3 H CH2CH3Br Br slow + Br δ+ δ– Step 2: C CH CH2CH3 H CH2CH3Br + Br C CH CH2CH3 Br H Br fast CH2CH3

-4- 5(d) • In step 1, both cis-hex-3-ene and trans-hex-3-ene react with HBr to give the same carbocation . This carbocation is trigonal planar with respect to the positively charged carbon. • In step 2, the Br− ion can attack the positively charged carbon of the carbocation from either side of the trigonal plane (via (a) or (b)) with equal probability. • Hence the product is a racemic mixture of the two enantiomers of 3 -bromohexane in a 1:1 ratio. 6 I Cl I Cl + Electrophilic Addition Step 1: I Cl I + δ+ δ– Cl+ slow more stable carbocation Step 2: I Cl+ fast Cl I 7(a) The reaction between ethene and hydrogen halide (HX) takes place via the electrophilic addition mechanism. In this case, the rate-determining step involves the cleavage of the hydrogen- halogen (H–X) covalent bond. The stronger the H–X covalent bond, the slower is the rate of the reaction and the lower is the reactivity of HX. Since the H–X bond strength decreases from HCl to HBr to HI, the rate of the reaction and hence reactivity increases from HCl to HBr to HI.

-5- 7(b) When 1-ethylcyclohexene reacts with HBr, electrophilic addition occurs. In the first step of the mechanism, two carbocations can be formed. + Br H δ+ δ– slow (a) (b) + Br + Br carbocation (1) carbocation (2) Carbocation (1) is a 3 o carbocation and its positive charge is dispersed by three electron - donating alkyl groups. This carbocation is more stable than carbocation (2), a 2o carbocation with its positive charge being dispersed by only two electron-donating alkyl groups. Being more stable, carbocation (1) is formed faster and is more available to participate in the second step of the reaction. Consequently, 1-bromo-1-ethylcyclohexane is the major product since it is formed faster from the more stable carbocation (1). + Br carbocation (1) fast Br 1-bromo-1-ethylcyclohexane (major product) 8 Answer: A (only 2-bromoethanol and 1-bromo-2-chloroethane are possible products) When ethene is shaken with Br2(aq) or shaken with Br2(aq) containing NaCl, the mechanism involved in the reactions in each case is electrophilic addition. The same first step occurs in each case to give the same carbocation since the only electrophile present is Br2: The second step involves the reaction between the bromine- containing carbocation and any nucleophile that is present. In Br2(aq), the nucleophiles present are H 2O and Br −, which react to give 2- bromoethanol and 1,2-dibromoethane respectively.

-6- In Br2(aq) containing NaCl, apart from H2O and Br−, Cl− is also present as a nucleophile and it reacts to give the additional product, 1-bromo-2-chloroethane. However, the reaction mixture does not contain Cl2. There is no electrophilic δ+Cl present to react with ethene and a chlorine-containing carbocation is not formed to react with Cl-. Hence 1,2-dichlorethane is not formed. Similarly, in the absence of the chlorine-containing carbocation, 2-chloroethanol is not formed. 9 hexa−1,5−diene trans-hexa-1,4-diene cis-hexa-1,4-diene A B C Deductions for Question 9 (fo r reference only) Compound A Evidence / Information Deduction and explanation A is a structural isomer of cyclohexene. A has the molecular formula, C6H10. in CCl4 darkC6H10 + 2Br 2 product A Electrophilic addition occurs. A is a diene or A has two C=C bonds per molecule. A butanedioic acid Oxidative cleavage of C=C bond occurs. The six-carbon A becomes a four-carbon acid. There is a loss of 2 carbon atoms as CO2 gas. Hence A contains two =CH 2 structural units (i.e. two terminal alkene units). Since the four-carbon acid is HO 2C(CH2)2CO2H, A has to be hexa−1,5−diene A does not display any stereoisomerism. Equations:

-7- in CCl4 dark + 2Br2 hexa−1,5−diene Br Br Br Br KMnO4(aq), H2SO4(aq) heat + 10[O] hexa−1,5−diene O O + 2CO2 + 2H 2O butanedioic acid OH OH Compound B Evidence / Information Deduction and explanation B is a positional isomer of A. B has the